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d296defa043a641f71ebe6408e5de095 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_21 | Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$
$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$ | Let $P,Q,R,X,Y,$ and $Z$ be the intersections $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},\overleftrightarrow{CD}\cap\overleftrightarrow{EF},\overleftrightarrow{EF}\cap\overleftrightarrow{AB},\overleftrightarrow{BC}\cap\overleftrightarrow{DE},\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively.
The sum of the interior angles of any hexagon is $720^\circ.$ Since hexagon $ABCDEF$ is equiangular, each of its interior angles is $720^\circ\div6=120^\circ.$ By angle chasing, we conclude that the interior angles of $\triangle PBC,\triangle QDE,\triangle RFA,\triangle XCD,\triangle YEF,$ and $\triangle ZAB$ are all $60^\circ.$ Therefore, these triangles are all equilateral triangles, from which $\triangle PQR$ and $\triangle XYZ$ are both equilateral triangles.
We are given that \begin{alignat*}{8} [PQR]&=\frac{\sqrt{3}}{4}\cdot PQ^2&&=192\sqrt3, \\ [XYZ]&=\frac{\sqrt{3}}{4}\cdot YZ^2&&=324\sqrt3, \end{alignat*} so we get $PQ=16\sqrt3$ and $YZ=36,$ respectively.
By equilateral triangles and segment addition, we find the perimeter of hexagon $ABCDEF:$ \begin{align*} AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ &=(YF+FA+AZ)+(PC+CD+DQ) \\ &=YZ+PQ \\ &=36+16\sqrt{3}. \end{align*} Finally, the answer is $36+16+3=\boxed{55}.$ | C | 55 |
d296defa043a641f71ebe6408e5de095 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_21 | Let $ABCDEF$ be an equiangular hexagon. The lines $AB, CD,$ and $EF$ determine a triangle with area $192\sqrt{3}$ , and the lines $BC, DE,$ and $FA$ determine a triangle with area $324\sqrt{3}$ . The perimeter of hexagon $ABCDEF$ can be expressed as $m +n\sqrt{p}$ , where $m, n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m + n + p$
$\textbf{(A)} ~47\qquad\textbf{(B)} ~52\qquad\textbf{(C)} ~55\qquad\textbf{(D)} ~58\qquad\textbf{(E)} ~63$ | Let the length $AB=x, BC=y.$ Then, we have \begin{align*} (y+2x)^2\cdot\frac{\sqrt 3}{4}&=324\sqrt3, \\ (x+2y)^2\cdot\frac{\sqrt 3}{4}&=192\sqrt3. \end{align*} We get \begin{align*} y+2x&=36, \\ x+2y&=16\sqrt3. \end{align*} We want $3x+3y,$ and it follows that \[3x+3y=(y+2x)+(x+2y)=36+16\sqrt3.\] Finally, the answer is $36+16+3=\boxed{55}.$ | C | 55 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$ | Suppose the roommate took sheets $a$ through $b$ , or equivalently, page numbers $2a-1$ through $2b$ . Because there are $(2b-2a+2)$ numbers taken, \[\frac{(2a-1+2b)(2b-2a+2)}{2}+19(50-(2b-2a+2))=\frac{50\cdot51}{2} \implies (2a+2b-39)(b-a+1)=\frac{50\cdot13}{2}=25\cdot13.\] The first possible solution that comes to mind is if $2a+2b-39=25, b-a+1=13 \implies a+b=32, b-a=12$ , which indeed works, giving $b=22$ and $a=10$ . The answer is $22-10+1=\boxed{13}$ | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$ | Suppose the smallest page number borrowed is $k,$ and $n$ pages are borrowed. It follows that the largest page number borrowed is $k+n-1.$
We have the following preconditions:
Together, we have \begin{align*} \frac{1275-\frac{(2k+n-1)n}{2}}{50-n}&=19 \\ 1275-\frac{(2k+n-1)n}{2}&=19(50-n) \\ 2550-(2k+n-1)n&=38(50-n) \\ 2550-(2k+n-1)n&=1900-38n \\ 650&=(2k+n-39)n. \end{align*} The factors of $650$ are \[1,2,5,10,13,25,26,50,65,130,325,650.\] Since $n$ is even, we only have a few cases to consider: \[\begin{array}{c|c|c} & & \\ [-2.25ex] \boldsymbol{n} & \boldsymbol{2k+n-39} & \boldsymbol{k} \\ [0.5ex] \hline & & \\ [-2ex] 2 & 325 & 181 \\ 10 & 65 & 47 \\ 26 & 25 & 19 \\ 50 & 13 & 1 \\ 130 & 5 & -43 \\ 650 & 1 & -305 \\ \end{array}\] Since $1\leq k \leq 49,$ only $k=47,19,1$ are possible:
Therefore, the only possibility is $k=19.$ We conclude that $n=26$ pages, or $\frac n2=\boxed{13}$ sheets, are borrowed. | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$ | Let $n$ be the number of sheets borrowed, with an average page number $k+25.5$ . The remaining $25-n$ sheets have an average page number of $19$ which is less than $25.5$ , the average page number of all $50$ pages, therefore $k>0$ . Since the borrowed sheets start with an odd page number and end with an even page number we have $k \in \mathbb N$ . We notice that $n < 25$ and $k \le (49+50)/2-25.5=24<25$
The weighted increase of average page number from $25.5$ to $k+25.5$ should be equal to the weighted decrease of average page number from $25.5$ to $19$ , where the weights are the page number in each group (borrowed vs. remained), therefore
\[2nk=2(25-n)(25.5-19)=13(25-n) \implies 13 | n \text{ or } 13 | k\]
Since $n, k < 25$ we have either $n=13$ or $k=13$ . If $n=13$ then $k=6$ . If $k=13$ then $2n=25-n$ which is impossible. Therefore the answer should be $n=\boxed{13}$ | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$ | Let $(2k-1)-2n$ be pages be borrowed, the sum of the page numbers on those pages is $(2n+2k+1)(n-k)$ while the sum of the rest pages is $1275-(2n+2k+1)(n-k)$ and we know the average of the rest is $\frac{1275-(2n+2k+1)}{50-2n+2k}$ which equals to $19$ ; multiply this out we got $950-38(n-k)=1275-(2n+2k+1)(n-k)$ and we got $(2n+2k-37)(n-k)=325$ . As $325=25\cdot13$ , we can see $n-k=13$ and that is desired $\boxed{13}$ | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$ | Let $c$ be the number of consecutive sheets Hiram’s roommate borrows, and let $b$ be the number of sheets preceding the $c$ borrowed sheets (i.e. if the friend borrows sheets $3$ $4$ , and $5$ , then $c=3$ and $b=2$ ).
The sum of the page numbers up till $b$ sheets is $1+2+3+\cdots + 2b=\frac{2b\cdot(2b+1)}{2} = b(2b+1)$ .
The last page number of the borrowed sheets would be $2(b+c)$ . Therefore, the sum of the remaining page numbers of the sheets after the $c$ borrowed sheets would be $2(b+c)+1 + 2(b+c)+2+\cdots+50$
The total number of page numbers after the borrow would be $50-2c$
Thus the average of the page numbers after the borrow would be: \[\frac{b(2b+1)+ 2(b+c)+1 + 2(b+c)+2+\cdots+50}{50-2c} =19.\] By the arithmetic series formula, this turns out to be: \[\frac{b(2b+1)+ \frac{(2(b+c)+1+50)\cdot(50-2c-2b)}{2}}{50-2c} =19\] because in the changed sum, there are $50$ numbers minus $2c$ borrowed numbers and $2b$ numbers from the first $b$ sheets.
This simplifies to \[\frac{b(2b+1)+ (2(b+c)+51)\cdot(25-c-b)}{50-2c} =19 \implies 19(50-2c)= b(2b+1)+ (2b+2c+51)\cdot(25-c-b).\] Noticing that some terms will cancel, we expand, leading to: \[950-38c=2b^2+b+50b-2bc-2b^2+50c-2c^2-2bc+1275-51c-51b\] \[\implies 950-38c=1275-4bc-c-2c^2 \implies 2c^2+4b-37c=325.\] Factoring, we get \[c(2c+4b-37)=325.\] The prime factorization of 325 is $5^2\cdot13$ . Recall that $0\leq c \leq 25$ , so $c$ could be $1$ $5$ $13$ , or $25$
We can rule out $c=25$ since Hiram would have no paper left over, so the average of the page numbers he has would be $0$ . We can now plug in the other answers for $c$ and we if we get a valid answer for $b$
Finally, just to make sure, we test $c=13 \implies 26+4b-37=25 \implies 4b=36 \implies b=9$
$b$ is an integer and $b+c=22$ , so everything checks out. The number of consecutive sheets borrowed by Hiram’s friend is $c=\boxed{13}$ | B | 13 |
e368d56db7592415d65f3334dc33d67e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 | Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2$ , the second sheet contains pages $3$ and $4$ , and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19$ . How many sheets were borrowed?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~13\qquad\textbf{(C)} ~15\qquad\textbf{(D)} ~17\qquad\textbf{(E)} ~20$ | The sum of all the page numbers is \[1+2+3+\cdots+50 = 1275.\] If we add the page numbers on each sheet, we get this sequence: \[3, 7, 11, \ldots, 99.\] So we can write the sum of the numbers on the first sheet that the roommate borrowed as $4n+3$ for some nonnegative integer, $n$ . If the roommate borrowed $k$ sheets, he borrowed sheets \[4n+3, 4n+7, \ldots, 4n+4k-1.\] The sum of the numbers in this sequence is \[\frac{k(8n+4k+2)}{2} = 4nk+2k^2+k.\] Since there are $2$ pages per sheet, there are $50-2k$ pages remaining, so the average page number of the remaining sheets is \[\frac{1275 - (4nk+2k^2+k)}{50-2k}.\] Therefore, \[\frac{1275 - (4nk+2k^2+k)}{50-2k} = 19,\] which simplifies to \[2k^2-37k-325 = -4nk.\] Factoring the left-hand side, \[(2k+13)(k-25)=-4nk.\] Since the right-hand side of this equation is divisible by $k$ , the left-hand side must also be divisible by $k$
In order for $(2k+13)(k-25)$ to be divisible by $k$ , either $2k+13$ or $k-25$ must be divisible by $k$ \[\frac{2k+13}{k}=2+\frac{13}{k},\] so $2k+13$ is divisible by $k$ only if $k$ is a factor of $13$ \[\frac{k-25}{k}=1-\frac{25}{k},\] so $k-25$ is divisible by $k$ only if $k$ is a factor of $25$
None of the answer choices are factors of $25$ , but answer choice B is a factor of $13$ . Hence, the answer is $\boxed{13}$ | B | 13 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid.
WLOG assume that A is in the center. \[\begin{tabular}{ c c c } ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{tabular}\] In this configuration, there are two cases, either all the A's lie on the same diagonal: \[\begin{tabular}{ c c c } ? & ? & A \\ ? & A & ? \\ A & ? & ? \end{tabular}\] or all the other two A's are on adjacent corners: \[\begin{tabular}{ c c c } A & ? & A \\ ? & A & ? \\ ? & ? & ? \end{tabular}\] In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.
In each case there is only one way to put the three B's and the three C's as shown in the diagrams. \[\begin{tabular}{ c c c } C & B & A \\ B & A & C \\ A & C & B \end{tabular}\] \[\begin{tabular}{ c c c } A & B & A \\ C & A & C \\ B & C & B \end{tabular}\] This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are $6$ ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{36}$ | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:
Case (1): The top-center and center-left chips have different colors. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, green); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); [/asy] There are three subcases for Case (1): [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, green); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, green); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, green); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green); fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, blue); fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red); fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue); fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue); fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, green); fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green); fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, blue); fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, red); fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, blue); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, green); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5)); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); [/asy] As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (1) has $3\cdot6=18$ ways in total.
Case (2): The top-center and center-left chips have the same color. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, blue); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); [/asy] There are three subcases for Case (2): [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, green); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue); fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, blue); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, green); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green); fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, red); fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red); fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue); fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue); fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, blue); fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, green); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green); fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, red); fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, blue); fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, green); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5)); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); [/asy] As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (2) has $3\cdot6=18$ ways in total.
Answer
Together, the answer is $18+18=\boxed{36}.$ | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | We consider all possible configurations of the red chips for which rotations matter: [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, red); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red); fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, red); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red); fill((18,2)--(19,2)--(19,3)--(18,3)--cycle, red); fill((20,1)--(21,1)--(21,2)--(20,2)--cycle, red); fill((19,0)--(20,0)--(20,1)--(19,1)--cycle, red); fill((24,1)--(25,1)--(25,2)--(24,2)--cycle, red); fill((26,1)--(27,1)--(27,2)--(26,2)--cycle, red); fill((25,2)--(26,2)--(26,3)--(25,3)--cycle, red); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5)); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle, linewidth(1.5)); draw((18,1)--(21,1), linewidth(1.5)); draw((18,2)--(21,2), linewidth(1.5)); draw((19,0)--(19,3), linewidth(1.5)); draw((20,0)--(20,3), linewidth(1.5)); draw((24,0)--(27,0)--(27,3)--(24,3)--cycle, linewidth(1.5)); draw((24,1)--(27,1), linewidth(1.5)); draw((24,2)--(27,2), linewidth(1.5)); draw((25,0)--(25,3), linewidth(1.5)); draw((26,0)--(26,3), linewidth(1.5)); label("Rotational",(1.5,4.5)); label("Symmetry",(1.5,3.75)); label("$2$ Configurations",(1.5,-0.75)); label("$4$ Configurations",(7.5,-0.75)); label("$4$ Configurations",(13.5,-0.75)); label("$4$ Configurations",(19.5,-0.75)); label("$4$ Configurations",(25.5,-0.75)); [/asy] As there are $2!=2$ permutations of blue and green for each configuration, the answer is $2\cdot(2+4+4+4+4)=\boxed{36}.$ | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | $(1) \quad$ $\begin{tabular}{ c c c } R & G & ? \\ B & R & ? \\ ? & ? & ? \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } R & G & B \\ B & R & G \\ R & G & B \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } R & G & R \\ B & R & B \\ G & B & G \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$
There are $3$ choices for $R$ $2$ choices for $G$ $R$ on the down left corner can be switched with $B$ on the upper right corner.
$(2) \quad$ $\begin{tabular}{ c c c } R & G & ? \\ B & R & ? \\ ? & ? & ? \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } R & G & B \\ B & R & G \\ G & B & R \end{tabular}$ $\quad 3 \cdot 2 = 6$
There are $3$ choices for $R$ $2$ choices for $G$
$(3) \quad$ $\begin{tabular}{ c c c } G & R & ? \\ R & B & ? \\ ? & ? & ? \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } G & R & B \\ R & B & G \\ G & R & B \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } G & R & G \\ R & B & R \\ B & G & B \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$
Note that $(3)$ is a $180$ ° rotation of $(1)$
$(4) \quad$ $\begin{tabular}{ c c c } G & R & ? \\ R & B & ? \\ ? & ? & ? \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c } G & R & B \\ R & B & G \\ B & G & R \end{tabular}$ $\quad 3 \cdot 2 = 6$
Note that $(4)$ is a $90$ ° rotation of $(2)$
Therefore, the answer is $2 \cdot (12 + 6) = \boxed{36}$ | E | 36 |
8b454e4062e3290a849d4bccf9f6438d | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | Case (1) : We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$ , so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are $2$ possible permutations for the last row. Thus, there are $3!\cdot2\cdot2=24$ possibilities.
Case (2) : All of the rows have two chips that are the same color and one that is different. There are obviously $3$ possible configurations for the first row, $2$ for the second, and $2$ for the third. Thus, there are $3\cdot2\cdot2=12$ possibilities.
Therefore, our answer is $24+12=\boxed{36}.$ | E | 36 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | Since $3\pi\approx9.42$ , we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{19}$ | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | $|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$ . Since $\pi$ is approximately $3.14$ $3\pi$ is approximately $9.42$ . We are trying to solve for $-9.42<x<9.42$ , where $x\in\mathbb{Z}$ . Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$ , for $x\in\mathbb{Z}$ . The number of integer values of $x$ is $9-(-9)+1=19$ . Therefore, the answer is $\boxed{19}$ | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | $3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$ . Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$ . Since $x$ is an integer, we must have $x$ between $-9$ and $9$ . There are a total of \[9-(-9) + 1 = \boxed{19}.\] PureSwag | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | Looking at the problem, we see that instead of directly saying $x$ , we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions: $9+9=18.$ But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{19}.$ | D | 19 |
e8d4559d876804950a8ea574d00a12b1 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_1 | How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$ | There are an odd number of integer solutions $x$ to this inequality since if any non-zero integer $x$ satisfies this inequality, then so does $-x,$ and we must also account for $0,$ which gives us the desired. Then, the answer is either $\textbf{(A)}$ or $\textbf{(D)},$ and since $3 \pi > 3 \cdot 3 > 9,$ the answer is at least $9 \cdot 2 + 1 = 19,$ yielding $\boxed{19}.$ | D | 19 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$ | Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations \[5j=2s\] \[j+s=28,\] and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is \[\boxed{8}.\] | C | 8 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$ | We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$ . Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{8}$ . ~samrocksnature | C | 8 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$ | Since there are an equal number of juniors and seniors on the debate team, suppose there are $x$ juniors and $x$ seniors. This number represents $25\% =\frac{1}{4}$ of the juniors and $10\%= \frac{1}{10}$ of the seniors, which tells us that there are $4x$ juniors and $10x$ seniors. There are $28$ juniors and seniors in the program altogether, so we get \[10x+4x=28,\] \[14x=28,\] \[x=2.\] Which means there are $4x=8$ juniors on the debate team, $\boxed{8}$ | C | 8 |
eb757777c1d63fc70bccd6269e66c29a | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_3 | In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$ | The amount of juniors must be a multiple of $4$ , since exactly $\frac{1}{4}$ of the students are on the debate team. Thus, we can immediately see that $\boxed{8}$ | C | 8 |
baf43db6a59c5d1245b61ddc16d9f589 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_4 | At a math contest, $57$ students are wearing blue shirts, and another $75$ students are wearing yellow shirts. The $132$ students are assigned into $66$ pairs. In exactly $23$ of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?
$\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64$ | There are $46$ students paired with a blue partner. The other $11$ students wearing blue shirts must each be paired with a partner wearing a shirt of the opposite color. There are $64$ students remaining. Therefore the requested number of pairs is $\tfrac{64}{2}=\boxed{32}$ ~Punxsutawney Phil | B | 32 |
5466a87961a99ab15febde232a84df47 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_5 | The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give $24$ , while the other two multiply to $30$ . What is the sum of the ages of Jonie's four cousins?
$\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25$ | First look at the two cousins' ages that multiply to $24$ . Since the ages must be single-digit, the ages must either be $3 \text{ and } 8$ or $4 \text{ and } 6.$
Next, look at the two cousins' ages that multiply to $30$ . Since the ages must be single-digit, the only ages that work are $5 \text{ and } 6.$ Remembering that all the ages must all be distinct, the only solution that works is when the ages are $3, 8$ and $5, 6$
We are required to find the sum of the ages, which is \[3 + 8 + 5 + 6 = \boxed{22}.\] | B | 22 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$ | Let there be $3x$ students in the morning class and $4x$ students in the afternoon class. The total number of students is $3x + 4x = 7x$ . The average is $\frac{3x\cdot84 + 4x\cdot70}{7x}=76$ . Therefore, the answer is $\boxed{76}$ | C | 76 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$ | Suppose the morning class has $m$ students and the afternoon class has $a$ students. We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{\# of Students} & \textbf{Mean} & \textbf{Total} \\ \hline & & & \\ [-2.5ex] \textbf{Morning} & m & 84 & 84m \\ \hline & & & \\ [-2.5ex] \textbf{Afternoon} & a & 70 & 70a \end{array}\] We are also given that $\frac ma=\frac34,$ which rearranges as $m=\frac34a.$
The mean of the scores of all the students is \[\frac{84m+70a}{m+a}=\frac{84\left(\frac34a\right)+70a}{\frac34a+a}=\frac{133a}{\frac74a}=133\cdot\frac47=\boxed{76}.\] ~MRENTHUSIASM | C | 76 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$ | Of the average, $\frac{3}{3+4}=\frac{3}{7}$ of the scores came from the morning class and $\frac{4}{7}$ came from the afternoon class. The average is $\frac{3}{7}\cdot 84+\frac{4}{7}\cdot 70=\boxed{76}.$ | C | 76 |
4572c0044f3f835d660400ecc5928044 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 | Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is $84$ , and the afternoon class's mean score is $70$ . The ratio of the number of students in the morning class to the number of students in the afternoon class is $\frac{3}{4}$ . What is the mean of the scores of all the students?
$\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78$ | WLOG, assume there are $3$ students in the morning class and $4$ in the afternoon class. Then the average is $\frac{3\cdot 84 + 4\cdot 70}{7}=\boxed{76}.$ | C | 76 |
b829d9c71c7c89f24fa97941134e7dac | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_7 | In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$
$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$ | Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one:
The diagram below shows one possible configuration of the four circles: [asy] /* diagram made by samrocksnature, edited by MRENTHUSIASM */ pair A=(10,0); pair B=(-10,0); draw(A--B); filldraw(circle((0,7),7),yellow); filldraw(circle((0,-5),5),yellow); filldraw(circle((0,-3),3),white); filldraw(circle((0,-1),1),white); dot((0,0)); label("$A$",(0,0),(0,1.5)); label("$\ell$",(10,0),(1.5,0)); [/asy] Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{65}.$ | D | 65 |
4787a0177733cbd1e33350d998204a63 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top? [asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$ | In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $D$ and $E,$ respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction. [asy] /* Made by MRENTHUSIASM */ size(11.5cm); for (real i=7.5; i<=14.5; ++i) { fill((i+0.5,i+0.5)--(i-0.5,i+0.5)--(i-0.5,i-0.5)--(i+0.5,i-0.5)--cycle,yellow); } fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); label("$A$",(14.5,14.5)); label("$B$",(13.5,13.5)); label("$C$",(0.5,14.5)); label("$E$",(1.5,13.5)); label("$D$",(0.5,13.5)); add(grid(15,15,linewidth(1.25))); draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); [/asy] By observations, we proceed as follows: \begin{alignat*}{6} A=15^2=225, \ B=13^2=169 \quad &\implies \quad &C &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &D &= &&C-1 &= 210& \\ \quad &\implies \quad &E &= &&B-12 &= 157&. \end{alignat*} Therefore, the answer is $D+E=\boxed{367}.$ | A | 367 |
4787a0177733cbd1e33350d998204a63 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top? [asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$ | In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $C$ and $G,$ respectively. [asy] /* Made by MRENTHUSIASM */ size(11.5cm); fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); label("$A$",(14.5,14.5)); label("$B$",(0.5,14.5)); label("$C$",(0.5,13.5)); label("$D$",(0.5,0.5)); label("$E$",(14.5,0.5)); label("$F$",(14.5,13.5)); label("$G$",(1.5,13.5)); add(grid(15,15,linewidth(1.25))); draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); [/asy] By observations, we proceed as follows: \begin{alignat*}{6} A=15^2=225 \quad &\implies \quad &B &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &C &= &&B-1 &= 210& \\ \quad &\implies \quad &D &= &&C-13 &= 197& \\ \quad &\implies \quad &E &= &&D-14 &= 183& \\ \quad &\implies \quad &F &= &&E-13 &= 170& \\ \quad &\implies \quad &G &= &&F-13 &= 157&. \end{alignat*} Therefore, the answer is $C+G=\boxed{367}.$ | A | 367 |
4787a0177733cbd1e33350d998204a63 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top? [asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$ | From the full diagram below, the answer is $210+157=\boxed{367} [/asy] This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options. | A | 367 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$ | The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$
By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$
Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \boxed{7}$ | D | 7 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$ | Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$ .
A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$ , and then translating the origin back to the point $(1, 5)$ \[a+b\cdot{i} \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = 5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$ .
Hence, we have \[6-b+(a+4)i = -3+6i \implies b=9, a=2,\] from which $b-a = 9-2 = \boxed{7}$ | D | 7 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$ | Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ .
If we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ .
We know that two vectors are perpendicular if their dot product is equal to $0$ , and that both points are the same distance ( $\sqrt {17}$ ) from $(1,5)$
Therefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)
Solving, we simplify the second equation to $y=4x+1$ , and plug it into the first equation.
We obtain $(x-1)^2 + (4x-4)^2 = 17$ . We can simplify this to the quadratic $17x^2-34x=0$ .
When we factor out $17x$ , we find that $x = 2$ or $x = 0$ . However, $x$ cannot equal $0$ .
Therefore, $x = 2$ , and plugging this into the second equation gives us that $y = 9$ .
Since the point is $(9, 2)$ , we compute $9-2 = \boxed{7}$ | D | 7 |
ec7f9db16fded448446ffd680982c3e0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_9 | The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$ | Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$
Let the original point be $\langle x,y \rangle.$ From point $(1,5),$ we form the vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \[\langle -4,1 \rangle \cdot \langle x-1, y-5 \rangle = 0 \implies -4x+y-1= 0.\] Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \pmod 3$ so that brings our potential answers down to $\text{\textbf{(A)}}$ and $\text{\textbf{(C)}}.$ If $A = 1$ from $\text{\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\boxed{7}$ is the answer. | D | 7 |
7def323ed742d23088be16a2c3447700 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_10 | An inverted cone with base radius $12 \mathrm{cm}$ and height $18 \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6$ | The volume of a cone is $\frac{1}{3} \cdot\pi \cdot r^2 \cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi$
The volume of a cylinder is $\pi \cdot r^2 \cdot h$ so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdot h$
We can equate these two expressions because the water volume stays the same like this $24\cdot24\cdot\pi\cdot h = 6\cdot144\pi$ . We get $4h = 6$ and $h=\frac{6}{4}$
So the answer is $\boxed{1.5}.$ | A | 1.5 |
7def323ed742d23088be16a2c3447700 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_10 | An inverted cone with base radius $12 \mathrm{cm}$ and height $18 \mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24 \mathrm{cm}$ . What is the height in centimeters of the water in the cylinder?
$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6$ | The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has $\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$ . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 = 4$ ).
Therefore, the height is divided by $3$ and divided by $4$ , which is $18 \div 3 \div 4 = \boxed{1.5}.$ | A | 1.5 |
d597cbb2c0b40ad7fb8f0bc71fc68697 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_11 | Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?
$\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64$ | Let the side lengths of the rectangular pan be $m$ and $n$ . It follows that $(m-2)(n-2) = \frac{mn}{2}$ , since half of the brownie pieces are in the interior. This gives $2(m-2)(n-2) = mn \iff mn - 4m - 4n + 8 = 0$ . Adding 8 to both sides and applying Simon's Favorite Factoring Trick , we obtain $(m-4)(n-4) = 8$ . Since $m$ and $n$ are both positive, we obtain $(m, n) = (5, 12), (6, 8)$ (up to ordering). By inspection, $5 \cdot 12 = \boxed{60}$ maximizes the number of brownies. | D | 60 |
d597cbb2c0b40ad7fb8f0bc71fc68697 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_11 | Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?
$\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64$ | Obviously, no side of the rectangular pan can have less than $5$ brownies beside it. We let one side of the pan have $5$ brownies, and let the number of brownies on its adjacent side be $x$ . Therefore, $5x=2\cdot3(x-2)$ , and solving yields $x=12$ and there are $5\cdot12=60$ brownies in the pan. $64$ is the only choice larger than $60$ , but it cannot be the answer since the only way to fit $64$ brownies in a pan without letting a side of it have less than $5$ brownies beside it is by forming a square of $8$ brownies on each side, which does not meet the requirement. Thus the answer is $\boxed{60}$ | D | 60 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$ | We can start by setting up an equation to convert $\underline{32d}$ base $n$ to base 10. To convert this to base 10, it would be $3{n}^2+2n+d.$ Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get $3{n}^2+2n = 263-d$
We can also set up equations to convert $\underline{324}$ base $n$ and $\underline{11d1}$ base 6 to base 10. The equation to covert $\underline{324}$ base $n$ to base 10 is $3{n}^2+2n+4.$ The equation to convert $\underline{11d1}$ base 6 to base 10 is ${6}^3+{6}^2+6d+1.$
Simplify ${6}^3+{6}^2+6d+1$ so it becomes $6d+253.$ Setting the above equations equal to each other, we have \[3{n}^2+2n+4 = 6d+253.\] Subtracting 4 from both sides gets $3{n}^2+2n = 6d+249.$
We can then use equations \[3{n}^2+2n = 263-d\] \[3{n}^2+2n = 6d+249\] to solve for $d$ . Set $263-d$ equal to $6d+249$ and solve to find that $d=2$
Plug $d=2$ back into the equation $3{n}^2+2n = 263-d$ . Subtract 261 from both sides to get your final equation of $3{n}^2+2n-261 = 0.$ We solve using the quadratic formula to find that the solutions are $9$ and $-29/3.$ Because the base must be positive, $n=9.$
Adding 2 to 9 gets $\boxed{11}$ | B | 11 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$ | $32d$ is greater than $263$ when both are interpreted in base 10, so $n$ is less than $10$ . Some trial and error gives $n=9$ $263$ in base 9 is $322$ , so the answer is $9+2=\boxed{11}$ | B | 11 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$ | We have \[3n^2 + 2n + d = 263\] \[3n^2 + 2n + 4 = 6^3 + 6^2 + 6d + 1\] Subtracting the 2nd from the 1st equation we get \begin{align*} d-4 &= 263 - (216 + 36 + 6d + 1) \\ &= 263 - 253 - 6d \\ &= 10 - 6d \end{align*} Thus we have $d=2.$ Substituting into the first, we have $3n^2 + 2n + 2 = 263 \Rightarrow 3n^2 + 2n - 261 = 0.$ Factoring, we have $(n-9)(3n+29)=0.$ A digit cannot be negative, so we have $n=9.$ Thus, $d+n=2+9=\boxed{11}$ | B | 11 |
5f081a1d94fea86c2ca21aa82c4ce759 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_13 | Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$ | Find that $d=2$ using one of the methods above. Then we have that $3n^2 + 2n = 261$ . We know that $n$ is an integer, so we can solve the equation $n(2n+3)=261$ (this is guaranteed to have a solution if we did this correctly). The prime factorization of $261$ is $3^2 \cdot 29$ , so the corresponding factor pairs are $(1, 261)$ $(3,87)$ , and $(9,29)$ . If $n = 9$ , then the equation is true, so $n + d = 9 + 2 = \boxed{11}$ | B | 11 |
4258c79f8131051481eacc756d986a1b | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_14 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | [asy] size(8cm); pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9); draw(O--A--B); draw(O--R); draw(O--L); label("$A$", A, NE); label("$B$", B, N); label("$R$", R, NE); label("$L$", L, NE); label("$O$", O, S); label("$d$", O--A, W); label("$2d$", A--B, W); label("$r$", O--R, S); label("$r$", O--L, NW); dot(O); dot(A); dot(B); dot(R); dot(L); draw(circle((0, 0), sqrt(370))); draw(-R -- (R.x, -R.y)); draw((-R.x, R.y) -- R); draw((-L.x, L.y) -- L); [/asy]
Since two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is
\[2d + d = 3d\]
and the distance between each of the chords is just $2d$ . Let the radius of the circle be $r$ . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
By the Pythagorean theorem, we can create the following system of equations:
\[19^2 + d^2 = r^2\]
\[17^2 + (2d + d)^2 = r^2\]
Solving, we find $d = 3$ , so $2d = \boxed{6}$ | B | 6 |
4258c79f8131051481eacc756d986a1b | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_14 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | [asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label("$O$", O, N); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, SW); label("$F$", F, SE); label("$P$", P, SW); label("$Q$", Q, S); [/asy] If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\triangle OCD$ with cevian $\overleftrightarrow{OP}$ gives \[19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.\] This simplifies to $13718+\tfrac{19}{2}d^{2}=38r^{2}$ . Similarly, another round of Stewart's Theorem applied to $\triangle OEF$ with cevian $\overleftrightarrow{OQ}$ gives \[17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.\] This simplifies to $9826+\tfrac{153}{2}d^{2}=34r^{2}$ . Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations \begin{align*} 361+\tfrac{1}{4}d^{2} &= r^{2} \\ 289+\tfrac{9}{4}d^{2} &= r^{2} \\ \end{align*} By transitive, $361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}$ . Therefore $(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{6}.$ | B | 6 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We square $x+\frac{1}{x}=\sqrt5$ to get $x^2+2+\frac{1}{x^2}=5$ . We subtract 2 on both sides for $x^2+\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\frac{1}{x^4}=9$ so $x^4+\frac{1}{x^4}=7$ . We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x^4-7+\frac{1}{x^4})$ . Therefore because $x^4+\frac{1}{x^4}$ is 7, it is equal to $x^7(0)=\boxed{0}$ | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | Multiplying both sides by $x$ and using the quadratic formula, we get $\frac{\sqrt{5} \pm 1}{2}$ . We can assume that it is $\frac{\sqrt{5}+1}{2}$ , and notice that this is also a solution the equation $x^2-x-1=0$ , i.e. we have $x^2=x+1$ . Repeatedly using this on the given (you can also just note Fibonacci numbers), \begin{align*} (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\ &=(2x^9+x^8)-7x^7+x^3 \\ &=(3x^8+2x^7)-7x^7+x^3 \\ &=(3x^8-5x^7)+x^3 \\ &=(-2x^7+3x^6)+x^3 \\ &=(x^6-2x^5)+x^3 \\ &=(-x^5+x^4+x^3) \\ &=-x^3(x^2-x-1) = \boxed{0} | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We can immediately note that the exponents of $x^{11}-7x^7+x^3$ are an arithmetic sequence, so they are symmetric around the middle term. So, $x^{11}-7x^7+x^3 = x^7(x^4-7+\frac{1}{x^4})$ . We can see that since $x+\frac{1}{x} = \sqrt{5}$ $x^2+2+\frac{1}{x^2} = 5$ and therefore $x^2+\frac{1}{x^2} = 3$ . Continuing from here, we get $x^4+2+\frac{1}{x^4} = 9$ , so $x^4-7+\frac{1}{x^4} = 0$ . We don't even need to find what $x^7$ is! This is since $x^7\cdot0$ is evidently $\boxed{0}$ , which is our answer. | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We begin by multiplying $x+\frac{1}{x} = \sqrt{5}$ by $x$ , resulting in $x^2+1 = \sqrt{5}x$ . Now we see this equation: $x^{11}-7x^{7}+x^3$ . The terms all have $x^3$ in common, so we can factor that out, and what we're looking for becomes $x^3(x^8-7x^4+1)$ . Looking back to our original equation, we have $x^2+1 = \sqrt{5}x$ , which is equal to $x^2 = \sqrt{5}x-1$ . Using this, we can evaluate $x^4$ to be $5x^2-2\sqrt{5}x+1$ , and we see that there is another $x^2$ , so we put substitute it in again, resulting in $3\sqrt{5}x-4$ . Using the same way, we find that $x^8$ is $21\sqrt{5}x-29$ . We put this into $x^3(x^8-7x^4+1)$ , resulting in $x^3(0)$ , so the answer is $\boxed{0}$ | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | The equation we are given is $x+\tfrac{1}{x}=\sqrt{5}...$ Yuck. Fractions and radicals! We multiply both sides by $x,$ square, and re-arrange to get \[x^2+1=\sqrt{5}x \implies x^4+2x^2+1=5x^2 \implies x^4-3x^2+1=0.\] Now, let us consider the expression we wish to acquire. Factoring out $x^3,$ we have \[x^3\left(x^8-7x^4+1\right) = x^3\left(x^8+2x^4+1-9x^4\right).\] Then, we notice that $x^8+2x^4+1=\left(x^4+1\right)^2.$ Furthermore, \[x^4+1=3x^2 \implies \left(x^4+1\right)^2=x^8+2x^4+1=9x^4.\] Thus, our answer is \[x^3\left(9x^4-9x^4\right) = x^3 \cdot 0 = \boxed{0} ~ 0.\] ~peace09 | B | 0 |
8e9258b3ecb8562e1bf9caf0585e55d9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_15 | The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | Multiplying by x and solving, we get that $x = \frac{\sqrt{5} \pm 1}{2}.$ Note that whether or not we take $x = \frac{\sqrt{5} + 1}{2}$ or we take $\frac{\sqrt{5} - 1}{2},$ our answer has to be the same. Thus, we take $x = \frac{\sqrt{5} - 1}{2} \approx 0.62$ . Since this number is small, taking it to high powers like $11$ $7$ , and $3$ will make the number very close to $0$ , so the answer is $\boxed{0}.$ ~AtharvNaphade | B | 0 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$ | The divisibility rule of $15$ is that the number must be congruent to $0$ mod $3$ and congruent to $0$ mod $5$ . Being divisible by $5$ means that it must end with a $5$ or a $0$ . We can rule out the case when the number ends with a $0$ immediately because the only integer that is uphill and ends with a $0$ is $0$ which is not positive. So now we know that the number ends with a $5$ . Looking at the answer choices, the answer choices are all pretty small, so we can generate all of the numbers that are uphill and are divisible by $3$ . These numbers are $15, 45, 135, 345, 1245, 12345$ , or $\boxed{6}$ numbers. | C | 6 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$ | First, note how the number must end in either $5$ or $0$ in order to satisfying being divisible by $15$ . However, the number can't end in $0$ because it's not strictly greater than the previous digits. Thus, our number must end in $5$ . We do casework on the number of digits.
Case 1: $1$ digit. No numbers work, so $0$ numbers.
Case 2: $2$ digits. We have the numbers $15, 45,$ and $75$ , but $75$ isn't an uphill number, so $2$ numbers
Case 3: $3$ digits. We have the numbers $135, 345$ , so $2$ numbers.
Case 4: $4$ digits. We have the numbers $1235, 1245$ and $2345$ , but only $1245$ satisfies this condition, so $1$ number.
Case 5: $5$ digits. We have only $12345$ , so $1$ number.
Adding these up, we have $2+2+1+1=\boxed{6}$ | C | 6 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$ | Like solution 2, we can proceed by using casework. A number is divisible by $15$ if is divisible by $3$ and $5.$ In this case, the units digit must be $5,$ otherwise no number can be formed.
Case 1: sum of digits = 6
There is only one number, $15.$
Case 2: sum of digits = 9
There are two numbers: $45$ and $135.$
Case 3: sum of digits = 12
There are two numbers: $345$ and $1245.$
Case 4: sum of digits = 15
There is only one number, $12345.$
We can see that we have exhausted all cases, because in order to have a larger sum of digits, then a number greater than $5$ needs to be used, breaking the conditions of the problem. The answer is $\boxed{6}.$ | C | 6 |
3f95f652da7ef08c38e0688eb263cdf6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 | Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~8$ | An integer is divisible by $15$ if it is divisible by $3$ and $5$ . Divisibility by $5$ means ending in $0$ or $5$ , but since no digit is less than $0$ , the only uphill integer ending in $0$ could be $0$ , which is not positive. This means the integer must end in $5$
All uphill integers ending in $5$ are formed by picking any subset of the sequence $(1,2,3,4)$ of digits (keeping their order), then appending a $5$ . Divisibility by $3$ means the sum of the digits is a multiple of $3$ , so our choice of digits must add to $0$ modulo $3$
$5 \equiv -1 \pmod{3}$ , so the other digits we pick must add to $1$ modulo $3$ . Since $(1,2,3,4) \equiv (1,-1,0,1) \pmod{3}$ , we can pick either nothing, or one residue $1$ (from $1$ or $4$ ) and one residue $-1$ (from $2$ ), and we can then optionally add a residue $0$ (from $3$ ). This gives $(1+2\cdot1)\cdot2 = \boxed{6}$ possibilities. | C | 6 |
9971bd3158bbac13c80c951631a3e3de | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_17 | Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given $2$ cards out of a set of $10$ cards numbered $1,2,3, \dots,10.$ The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- $11,$ Oscar-- $4,$ Aditi-- $7,$ Tyrone-- $16,$ Kim-- $17.$ Which of the following statements is true?
$\textbf{(A) }\text{Ravon was given card 3.}$
$\textbf{(B) }\text{Aditi was given card 3.}$
$\textbf{(C) }\text{Ravon was given card 4.}$
$\textbf{(D) }\text{Aditi was given card 4.}$
$\textbf{(E) }\text{Tyrone was given card 7.}$ | By logical deduction, we consider the scores from lowest to highest: \begin{align*} \text{Oscar's score is 4.} &\implies \text{Oscar is given cards 1 and 3.} \\ &\implies \text{Aditi is given cards 2 and 5.} \\ &\implies \text{Ravon is given cards 4 and 7.} && (\bigstar) \\ &\implies \text{Tyrone is given cards 6 and 10.} \\ &\implies \text{Kim is given cards 8 and 9.} \end{align*} Therefore, the answer is $\boxed{4.}$ | C | 4. |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then returned to the set, the average value of the integers rises to $40$ . The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ . What is the average value of all the integers in the set $S$
$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$ | Let the lowest value be $L$ and the highest $G$ , and let the sum be $Z$ and the amount of numbers $n$ . We have $\frac{Z-G}{n-1}=32$ $\frac{Z-L-G}{n-2}=35$ $\frac{Z-L}{n-1}=40$ , and $G=L+72$ . Clearing denominators gives $Z-G=32n-32$ $Z-L-G=35n-70$ , and $Z-L=40n-40$ . We use $G=L+72$ to turn the first equation into $Z-L=32n+40$ . Since $Z-L=40(n-1)$ we substitute it into the equation which gives $n=10$ . Turning the second into $Z-2L=35n+2$ using $G=L+72$ we see $L=8$ and $Z=368$ so the average is $\frac{Z}{n}=\boxed{36.8}$ ~aop2014 | D | 36.8 |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then returned to the set, the average value of the integers rises to $40$ . The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ . What is the average value of all the integers in the set $S$
$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$ | Let $x$ be the greatest integer, $y$ be the smallest, $z$ be the sum of the numbers in S excluding $x$ and $y$ , and $k$ be the number of elements in S.
Then, $S=x+y+z$
First, when the greatest integer is removed, $\frac{S-x}{k-1}=32$
When the smallest integer is also removed, $\frac{S-x-y}{k-2}=35$
When the greatest integer is added back, $\frac{S-y}{k-1}=40$
We are given that $x=y+72$
After you substitute $x=y+72$ , you have 3 equations with 3 unknowns $S,$ $y$ and $k$
$S-y-72=32k-32$
$S-2y-72=35k-70$
$S-y=40k-40$
This can be easily solved to yield $k=10$ $y=8$ $S=368$
$\therefore$ average value of all integers in the set $=S/k = 368/10 = \boxed{36.8}$ | D | 36.8 |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then returned to the set, the average value of the integers rises to $40$ . The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ . What is the average value of all the integers in the set $S$
$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$ | We should plug in $36.2$ and assume everything is true except the $35$ part. We then calculate that part and end up with $35.75$ . We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$ . The answer is then $\boxed{36.8}$ . You can work backwards because it is multiple choice and you don't have to do critical thinking. ~Lopkiloinm | D | 36.8 |
0d72b274cc03a643f28139c6efc2fcf0 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_19 | Suppose that $S$ is a finite set of positive integers. If the greatest integer in $S$ is removed from $S$ , then the average value (arithmetic mean) of the integers remaining is $32$ . If the least integer in $S$ is also removed, then the average value of the integers remaining is $35$ . If the greatest integer is then returned to the set, the average value of the integers rises to $40$ . The greatest integer in the original set $S$ is $72$ greater than the least integer in $S$ . What is the average value of all the integers in the set $S$
$\textbf{(A) }36.2 \qquad \textbf{(B) }36.4 \qquad \textbf{(C) }36.6\qquad \textbf{(D) }36.8 \qquad \textbf{(E) }37$ | Let $S = \{a_1, a_2, a_3, \hdots, a_n\}$ with $a_1 < a_2 < a_3 < \hdots < a_n.$ We are given the following: \[{\begin{cases} \sum_{i=1}^{n-1} a_i = 32(n-1) = 32n-32, \\ \sum_{i=2}^n a_i = 40(n-1) = 40n-40, \\ \sum_{i=2}^{n-1} a_i = 35(n-2) = 35n-70, \\ a_n-a_1 = 72 \implies a_1 + 72 = a_n. \end{cases}}\] Subtracting the third equation from the sum of the first two, we find that \[\sum_{i=1}^n a_i = \left(32n-32\right) + \left(40n-40\right) - \left(35n-70\right) = 37n - 2.\] Furthermore, from the fourth equation, we have \[\sum_{i=2}^{n} a_i - \sum_{i=1}^{n-1} a_i = \left[\left(a_1 + 72\right) + \sum_{i=2}^{n-1} a_i\right] - \left[\left(a_1\right) + \sum_{i=2}^{n-1} a_i\right] = \left(40n-40\right)-\left(32n-32\right).\] Combining like terms and simplifying, we have \[72 = 8n-8 \implies 8n = 80 \implies n=10.\] Thus, the sum of the elements in $S$ is $37 \cdot 10 - 2 = 368,$ and since there are 10 elements in $S,$ the average of the elements in $S$ is $\tfrac{368}{10}=\boxed{36.8}$ | D | 36.8 |
4693b8f76efa4eac7ea2286add2491ed | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_20 | The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$ [asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]
$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$ | [asy] /* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-1.85,2); pair G=(-3.1,2); draw(A--G--A--F, lightgray); draw(B--F--C, lightgray); draw(E--G--D, lightgray); dot(F^^G, lightgray); draw(A--B--C--D--E--A); draw(A--C--A--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E); [/asy]
Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\sqrt{3}$ and $CD=2$ . Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$ , so the area is $\sqrt{11}$ . Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{23}$ | D | 23 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy] | We can set the point on $CD$ where the fold occurs as point $F$ . Then, we can set $FD$ as $x$ , and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$ , we get,
\[x^2 + \left(\frac{1}{3}\right)^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}\]
We know this is a 3-4-5 triangle because the side lengths are $\frac{3}{9}, \frac{4}{9}, \frac{5}{9}$ . We also know that $EAC'$ is similar to $C'DF$ because angle $EC'F$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C'DF \times \frac{AC'}{DF}$ . That's just $\frac{4}{3} \times \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} \times \frac{3}{2} = 2$ . Therefore, the final answer is $\boxed{2}$ | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy] | Let the line we're reflecting over be $\ell$ , and let the points where it hits $AB$ and $CD$ , be $M$ and $N$ , respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line $\ell$ . The segment $CC'$ has slope $\frac{0 - 1}{1 - 2/3} = -3$ , implying line $\ell$ has a slope of $\frac{1}{3}$ . Also, the midpoint of segment $CC'$ is $\left( \frac{5}{6}, \frac{1}{2} \right)$ , so line $\ell$ passes through this point. Then, we get the equation of line $\ell$ is simply $y = \frac{1}{3} x + \frac{2}{9}$ . Then, if the point where $B$ is reflected over line $\ell$ is $B'$ , then we get $BB'$ is the line $y = -3x$ . The intersection of $\ell$ and segment $BB'$ is $\left( - \frac{1}{15}, \frac{1}{5} \right)$ . So, we get $B' = \left(- \frac{2}{15}, \frac{2}{5} \right)$ . Then, line segment $B'C'$ has equation $y = \frac{3}{4} x + \frac{1}{2}$ , so the point $E$ is the $y$ -intercept, or $\left(0, \frac{1}{2} \right)$ . This implies that $AE = \frac{1}{2}, AC' = \frac{2}{3}$ , and by the Pythagorean Theorem, $EC' = \frac{5}{6}$ (or you could notice $\triangle AEC'$ is a $3-4-5$ right triangle). Then, the perimeter is $\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2$ , so our answer is $\boxed{2}$ . ~rocketsri | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy] | Assume that E is the midpoint of $\overline{AB}$ . Then, $\overline{AE}=\frac{1}{2}$ and since $C'D=\frac{1}{3}$ $\overline{AC'}=\frac{2}{3}$ . By the Pythagorean Theorem, $\overline{EC'}=\frac{5}{6}$ . It easily follows that our desired perimeter is $\boxed{2}$ ~samrocksnature | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy] | As described in Solution 1, we can find that $DF=\frac{4}{9}$ , and $C'F = \frac{5}{9}.$
Then, we can find we can find the length of $\overline{AE}$ by expressing the length of $\overline{EF}$ in two different ways, in terms of $AE$ . If let $AE = a$ , by the Pythagorean Theorem we have that $EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.$ Therefore, since we know that $\angle EC'F$ is right, by Pythagoras again we have that $EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}}\right)^2 + \left(\frac{5}{9}\right)^2} = \sqrt{a^2 + \frac{61}{81}}.$
Another way we can express $EF$ is by using Pythagoras on $\triangle XEF$ , where $X$ is the foot of the perpendicular from $F$ to $\overline{AE}.$ We see that $ADFX$ is a rectangle, so we know that $AD = 1 = FX$ . Secondly, since $FD = \frac{4}{9}, EX = a - \frac{4}{9}$ . Therefore, through the Pythagorean Theorem, we find that $EF = \sqrt{\left(a-\frac{4}{9}\right)^2 + 1^2} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.$
Since we have found two expressions for the same length, we have the equation $\sqrt{a^2 + \frac{61}{81}} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.$ Solving this, we find that $a=\frac{1}{2}$
Finally, we see that the perimeter of $\triangle AEC'$ is $\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},$ which we can simplify to be $2$ . Thus, the answer is $\boxed{2}.$ ~laffytaffy | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy] | Draw a perpendicular line from $\overline{AB}$ at $E$ , and let it intersect $\overline{DC}$ at $E'$ . The angle between $\overline{AB}$ and $\overline{EE'}$ is $2\theta$ , where $\theta$ is the angle between the fold and a line perpendicular to $\overline{AD}$ . The slope of the fold is $\frac{1}{3}$ because it is perpendicular to $\overline{CC'}$ $\overline{CC'}$ has a slope of $-3$ using points $C'$ and $C$ , and perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of $\overline{EC'}$ is $\frac{3}{4}$ , which implies $\overline{AE}$ $\frac{1}{2}$ . By the Pythagorean Theorem, $\overline{EC'}=\frac{5}{6}$ . It easily follows that our desired perimeter is $\boxed{2}$ ~forrestc | A | 2 |
a50f97bda47d3b2c8e4d2f80819b59d4 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_21 | A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy] | It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means: \[C' F + DF = CF + DF = CD = AD = 3 C'D \implies C' D : DF : C' F = 3 : 4 : 5.\]
For a similar triangle, the ratio of the perimeter to the larger leg is $\frac {3 + 4 +5}{4} = 3.$
$\triangle AEC' \sim \triangle DC'F \implies$ the perimeter of triangle $\bigtriangleup AEC'$ is $3 AC' = \boxed{2}.$ | A | 2 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$
$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$ | Let our denominator be $(5!)^3$ , so we consider all possible distributions.
We use PIE (Principle of Inclusion and Exclusion) to count the successful ones.
When we have at $1$ box with all $3$ balls the same color in that box, there are $_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3$ ways for the distributions to occur ( $_{5} C _{1}$ for selecting one of the five boxes for a uniform color, $_{5} P _{1}$ for choosing the color for that box, $4!$ for each of the three people to place their remaining items).
However, we overcounted those distributions where two boxes had uniform color, and there are $_{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3$ ways for the distributions to occur ( $_{5} C _{2}$ for selecting two of the five boxes for a uniform color, $_{5} P _{2}$ for choosing the color for those boxes, $3!$ for each of the three people to place their remaining items).
Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth.
Our success by PIE is \[_{5} C _{1} \cdot _{5} P _{1} \cdot (4!)^3 - _{5} C _{2} \cdot _{5} P _{2} \cdot (3!)^3 + _{5} C _{3} \cdot _{5} P _{3} \cdot (2!)^3 - _{5} C _{4} \cdot _{5} P _{4} \cdot (1!)^3 + _{5} C _{5} \cdot _{5} P _{5} \cdot (0!)^3 = 120 \cdot 2556.\] \[\frac{120 \cdot 2556}{120^3}=\frac{71}{400},\] yielding an answer of $\boxed{471}$ | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$
$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$ | As In Solution 1, the probability is \[\frac{\binom{5}{1}\cdot 5\cdot (4!)^3 - \binom{5}{2}\cdot 5\cdot 4\cdot (3!)^3 + \binom{5}{3}\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - \binom{5}{4}\cdot 5\cdot 4\cdot 3\cdot 2\cdot (1!)^3 + \binom{5}{5}\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{(5!)^3}\] \[= \frac{5\cdot 5\cdot (4!)^3 - 10\cdot 5\cdot 4\cdot (3!)^3 + 10\cdot 5\cdot 4\cdot 3\cdot (2!)^3 - 5\cdot 5! + 5!}{(5!)^3}.\] Dividing by $5!$ , we get \[\frac{5\cdot (4!)^2 - 10\cdot (3!)^2 + 10\cdot (2!)^2 - 5 + 1}{(5!)^2}.\] Dividing by $4$ , we get \[\frac{5\cdot 6\cdot 24 - 10\cdot 9 + 10 - 1}{30\cdot 120}.\] Dividing by $9$ , we get \[\frac{5\cdot 2\cdot 8 - 10 + 1}{10\cdot 40} = \frac{71}{400} \implies \boxed{471}\] | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$
$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$ | Use complementary counting.
Denote $T_n$ as the total number of ways to put $n$ colors to $n$ boxes by 3 people out of which $f_n$ ways are such that no box has uniform color. Notice $T_n = (n!)^3$ . From this setup we see the question is asking for $1-\frac{f_5}{(5!)^3}$ . To find $f_5$ we want to exclude the cases of a) one box of the same color, b) 2 boxes of the same color, c) 3 boxes of same color, d) 4 boxes of the same color, and e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have
\[f_5=T_5 -{\binom{5}{1}\binom{5}{1}\cdot f_4 - \binom{5}{2}\binom{5}{2}\cdot 2!\cdot f_3 - \binom{5}{3}\binom{5}{3}\cdot 3!\cdot f_2 - 5!}\]
In case b), there are $\binom{5}{2}$ ways to choose 2 boxes that have the same color, $\binom{5}{2}$ ways to choose the two colors, 2! ways to permute the 2 chosen colors, and $f_3$ ways so that the remaining 3 boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways to permute 5 colors is $5!$ . Now, we just need to calculate $f_2$ $f_3$ and $f_4$
We have $f_2=T_2-2 = (2!)^3 - 2 = 6$ , since we subtract the number of cases where the boxes contain uniform colors, which is 2.
In the same way, $f_3=T_3-\Big[3!+ \binom{3}{1}\binom{3}{1}\cdot f_2 \Big] = 156$ - again, we must subtract the number of ways at least 1 box has uniform color. There are $3!$ ways if 3 boxes each contains uniform color. Two boxes each contains uniform color is the same as previous. If one box has the same color, there are $\binom{3}{1}$ ways to pick a box, and $\binom{3}{1}$ ways to pick a color for that box, 1! ways to permute the chosen color, and $f_2$ ways for the remaining 2 boxes to have non-uniform colors. Similarly, $f_4=(4!)^3-\Big[ 4!+ \binom{4}{2}\binom{4}{2}\cdot 2! \cdot f_2+ \binom{4}{1}\binom{4}{1}\cdot f_3\Big] = 10,872$
Thus, $f_5 = f_5=(5!)^3-\Big[\binom{5}{1}\binom{5}{1}\cdot f_4+ \binom{5}{2}\binom{5}{2}\cdot 2!\cdot f_3+\binom{5}{3}\binom{5}{3}\cdot 3!\cdot f_2 + 5!\Big] = (5!)^3 - 306,720$
Thus, the probability is $\frac{306,720}{(5!)^3} = 71/400$ and the answer is $\boxed{471}$ | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$
$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$ | WLOG fix which block Ang places into each box. (This can also be thought of as labeling each box by the color of Ang's block.) There are then $(5!)^2$ total possibilities.
As in Solution 1 , we use PIE. With $1$ box of uniform color, there are ${}_{5} C _{1} \cdot (4!)^2$ possibilities ( ${}_{5} C _{1}$ for selecting one of the five boxes (whose color is fixed by Ang), $4!$ for each of Ben and Jasmin to place their remaining items). We overcounted cases with $2$ boxes, of which there are ${}_{5} C _{2} \cdot (3!)^2$ possibilities, and so on.
The probability is thus \[\frac{{}_{5} C _{1} \cdot (4!)^2 - {}_{5} C _{2} \cdot (3!)^2 + {}_{5} C _{3} \cdot (2!)^2 - {}_{5} C _{4} \cdot (1!)^2 + {}_{5} C _{5} \cdot (0!)^2}{(5!)^2}\] \[= \frac{5 \cdot (4!)^2 - 10 \cdot (3!)^2 + 10 \cdot (2!)^2 - 5 + 1}{(5!)^2}\] at which point we can proceed as in #Alternate simplification to simplify to $\frac{71}{400} \implies \boxed{471}$ | D | 471 |
f7f8e8d38178389110d6d8f6965bd7ad | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_22 | Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$
$\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542$ | $!n$ denotes the number of derangements of $n$ elements, i.e. the number of permutations where no element appears in its original position. Recall the recursive formula $!0 = 1, !1 = 0, !n = (n-1)(!(n-1)+{!}(n-2))$
We will consider the number of candidate boxes (ones that currently remain uniform-color) after each person places their blocks.
After Ang, all $5$ boxes are candidates, and each has a fixed color its blocks must be to remain uniform-color.
Ben has $5!$ total ways to place blocks. There are $\tbinom{5}{k}\cdot{!}k$ ways to disqualify $k$ candidates, $0 \le k \le 5$ . (There are $\tbinom{5}{k}$ ways to choose the candidates to disqualify, and $!k$ ways to arrange the disqualified candidates' same-color blocks.)
Jasmin has $5!$ total ways to place blocks. Currently, $k$ boxes are no longer candidates. Consider the number of ways for Jasmin to place blocks such that no box remains a candidate, where $n$ is the number of boxes and $k$ is the number of non-candidates, and call it $D(n,k)$ (not to be confused with partial derangements). We wish to find $5!-D(5,k)$ for each $k$ $0 \le k \le 5$
Notice that $D(n,0) = {!}n$ , since all boxes are candidates and no block can be placed in the same-colored box. Also notice the recursive formula $D(n,k) = D(n-1,k-1)+D(n,k-1)$ , since the first non-candidate box can either contain its same-color block (in which case it can be ignored), or a different-color block (in which case it can be treated as a candidate).
We can make a table ( $n$ horizontal, $k$ vertical): \[\begin{array}{r|rrrrrr} D(n,k) & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 1 & 0 & 1 & 2 & 9 & 44 \\ 1 & & 1 & 1 & 3 & 11 & 53 \\ 2 & & & 2 & 4 & 14 & 64 \\ 3 & & & & 6 & 18 & 78 \\ 4 & & & & & 24 & 96 \\ 5 & & & & & & 120 \\ \end{array}\] (Note that $D(n,n) = n!$ since there are no candidates to begin with, which checks out with the diagonal.)
The desired values of $D(5,k)$ are in the rightmost column.
The probability that at least one box is of uniform color is thus \begin{align*} & \frac{\sum_{k=0}^{5}{\left(\binom{5}{k}\cdot{!}k\right)(5!-D(5,k))}}{(5!)^2} \\ ={}& \frac{(1\cdot1)(120-44)+(5\cdot0)(120-53)+(10\cdot1)(120-64)+(10\cdot2)(120-78)+(5\cdot9)(120-96)+(1\cdot44)(120-120)}{(5!)^2} \\ ={}& \frac{76+0+560+840+1080+0}{(5!)^2} \\ ={}& \frac{19+140+210+270}{(5!)(5\cdot3\cdot2)} \\ ={}& \frac{639}{(5!)(5\cdot3\cdot2)} \\ ={}& \frac{71}{(5\cdot4\cdot2)(5\cdot2)} \\ ={}& \frac{71}{400} \implies \boxed{471} | D | 471 |
b1b6cc5abf05ac743aa96a4df5c90c24 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_23 | A square with side length $8$ is colored white except for $4$ black isosceles right triangular regions with legs of length $2$ in each corner of the square and a black diamond with side length $2\sqrt{2}$ in the center of the square, as shown in the diagram. A circular coin with diameter $1$ is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as $\frac{1}{196}\left(a+b\sqrt{2}+\pi\right)$ , where $a$ and $b$ are positive integers. What is $a+b$ [asy] /* Made by samrocksnature */ draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0)); fill((2,0)--(0,2)--(0,0)--cycle, black); fill((6,0)--(8,0)--(8,2)--cycle, black); fill((8,6)--(8,8)--(6,8)--cycle, black); fill((0,6)--(2,8)--(0,8)--cycle, black); fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black); filldraw(circle((2.6,3.31),0.5),gray); [/asy]
$\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72$ | To find the probability, we look at the $\frac{\text{success region}}{\text{total possible region}}$ . For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least $\frac{1}{2}$ , as it's the radius of the coin. This implies the $\text{total possible region}$ is a square with side length $8 - \frac{1}{2} - \frac{1}{2} = 7$ , with an area of $49$ . Now, we consider cases where needs to land to partially cover a black region.
Near The Center Square
We can have the center of the coin land within $\frac{1}{2}$ outside of the center square, or inside of the center square. So, we have a region with $\frac{1}{2}$ emanating from every point on the exterior of the square, forming four quarter circles and four rectangles. The four quarter circles combine to make a full circle of radius $\frac{1}{2}$ , so the area is $\frac{\pi}{4}$ . The area of a rectangle is $2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2$ , so $4$ of them combine to an area of $4 \sqrt 2$ . The area of the black square is simply $\left(2\sqrt 2\right)^2 = 8$ . So, for this case, we have a combined total of $8 + 4\sqrt 2 + \frac{\pi}{4}$ . Onto the second (and last) case.
Near A Triangle
We can also have the coin land within $\frac{1}{2}$ outside of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by $4$ . Consider the above diagram. We can draw an altitude from the bottom corner of the square to hit the hypotenuse of the green triangle. The length of this when passing through the black region is $\sqrt 2$ , and when passing through the white region (while being contained in the green triangle) is $\frac{1}{2}$ . However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of $\dfrac{1}{2}$ which is $\dfrac{\sqrt{2}}{2}.$ So, the altitude of the green triangle is $\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}$ . Then, recall, the area of an isosceles right triangle is $h^2$ , where $h$ is the altitude from the right angle. So, squaring this, we get $\frac{3 + 2\sqrt 2}{4}$ . Now, we have to multiply this by $4$ to account for all of the black triangles, to get $3 + 2\sqrt 2$ as the final area for this case.
Finishing
Then, to have the coin touching a black region, we add up the area of our successful regions, or $8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}$ . The total region is $49$ , so our probability is $\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}$ , which implies $a+b = 44+24 = 68$ . This corresponds to answer choice $\boxed{68}$ | C | 68 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$
[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?
$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$ | We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position.
First we note that symmetrical positions are P-positions, as the second player can win by mirroring the first player's moves. It follows that $(6, 1, 1)$ is an N-position, since we can win by moving to $(2, 2, 1, 1)$ ; this rules out $\textbf{(A)}$ . We next look at $(6, 2, 1)$ . The possible next states are \[(6, 2), \enskip (6, 1, 1), \enskip (6, 1), \enskip (5, 2, 1), \enskip (4, 2, 1, 1), \enskip (4, 2, 1), \enskip (3, 2, 2, 1), \enskip (3, 2, 1, 1), \enskip (2, 2, 2, 1).\] None of these are symmetrical, so we might reasonably suspect that they are all N-positions. Indeed, it just so happens that for all of these states except $(6, 2)$ and $(6, 1)$ , we can win by moving to $(2, 2, 1, 1)$ ; it remains to check that $(6, 2)$ and $(6, 1)$ are N-positions.
To save ourselves work, it would be nice if we could find a single P-position directly reachable from both $(6, 2)$ and $(6, 1)$ . We notice that $(3, 2, 1)$ is directly reachable from both states, so it would suffice to show that $(3, 2, 1)$ is a P-position. Indeed, the possible next states are \[(3, 2), \enskip (3, 1, 1), \enskip (3, 1), \enskip (2, 2, 1), \enskip (2, 1, 1, 1), \enskip (2, 1, 1),\] which allow for the following refutations:
\begin{align*} &(3, 2) \to (2, 2), && &&(3, 1, 1) \to (1, 1, 1, 1), && &&(3, 1) \to (1, 1), \\ &(2, 2, 1) \to (2, 2), && &&(2, 1, 1, 1) \to (1, 1, 1, 1), && &&(2, 1, 1) \to (1, 1). \end{align*}
Hence, $(3, 2, 1)$ is a P-position, so $(6, 2)$ and $(6, 1)$ are both N-positions, along with all other possible next states from $(6, 2, 1)$ as noted before. Thus, $(6, 2, 1)$ is a P-position, so our answer is $\boxed{6,2,1}$ as in Solution 2.) | B | 6,2,1 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$
[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?
$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$ | $(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.
$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.
$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.
$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun, which is symmetric, so Beth will lose.
That leaves $(6,2,1)$ or $\boxed{6,2,1}$ | B | 6,2,1 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$
[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?
$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$ | Consider the following, much simpler game.
Arjun and Beth can each either take 1 or 2 bricks from the right-hand-side of a continuous row of initially $n$ bricks.
It is easy to see that for $n$ a multiple of 3, Beth can "mirror" whatever Arjun plays: if he takes 1, she takes 2, and if he takes 2, she takes 1. With this strategy, Beth always takes the last brick. If $n$ is not a multiple of 3, then Arjun takes whichever amount puts Beth in the losing position.
The total number of bricks in the initial states given by the answer choices is $8, 9, 10, 10, 11$ . Thus, answer choice $\textbf{(B)}$ appears promising as a winning position for Beth. The difference between this game and the simplified game is that in certain positions, namely those consisting of fragments of size only 1, taking 2 bricks is not allowed. We can assume that for the starting position $(6, 2, 1)$ , Beth always has a move to ensure that she can continue to mirror Arjun throughout the game. (This could be proven rigorously with lots of casework. In particular, she must avoid providing Arjun with a position of 3 continuous bricks, because then he could take the middle block and force a win.) The assumption seems reasonable, so the answer is $\boxed{6,2,1}$ | B | 6,2,1 |
ceaf997432282068e57a84da83652abc | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_24 | Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$
[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?
$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$ | We can start by guessing and checking our solutions. Let's start with $A$ $(6,1,1)$ . Arjun goes first, and he has a winning strategy. His strategy is to cut the 6 block in half, so whatever Beth does, Arjun will copy. If we see this in practice, Arjun has made the block in to $(2,2,1,1)$ . If Beth takes 1, he will take one. If Beth takes 2, he will take 2 as well.
Then, we can guess solution $B$ $(6,2,1)$ . If Arjun starts by taking 1 of the blocks, he will either have $(6,1,1)$ $(6,2)$ $(x,y,2,1)$ (Where $x+y = 5$ ). If he has $(6,1,1)$ , Beth can take 2 out of the middle of $(6,1,1)$ to get $(2,2,1,1)$ which we know Arjun will lose because we know by symmetry from answer choice $A$ . If he has $(6,2)$ , Beth will take out one from the edge, and we get $(5,2)$ . Beth can now copy whatever Arjun does, so she will win.
Next, if he takes one out to get $(x,y,2,1)$ , we can either have $(4,1,2,1)$ or $(3,2,2,1)$ . If it is the first case, Beth can take the edge 2 out of the 4, which gives us $(2,1,2,1)$ and whatever Arjun does, Beth can do, so she has established a win. If it is the 2nd case, Beth can take out 2 from the 3 to get $(1,2,2,1)$ , in which, Beth can copy whatever Arjun does, so Beth will win. So in all 4 cases, Beth wins. So our answer is $\boxed{6,2,1}$ | B | 6,2,1 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | I know that I want about $\frac{2}{3}$ of the box of integer coordinates above my line. There are a total of 30 integer coordinates in the desired range for each axis which gives a total of 900 lattice points. I estimate that the slope, m, is $\frac{2}{3}$ . Now, although there is probably an easier solution, I would try to count the number of points above the line to see if there are 600 points above the line. The line $y=\frac{2}{3}x$ separates the area inside the box so that $\frac{2}{3}$ of the are is above the line.
I find that the number of coordinates with $x=1$ above the line is 30, and the number of coordinates with $x=2$ above the line is 29. Every time the line $y=\frac{2}{3}x$ hits a y-value with an integer coordinate, the number of points above the line decreases by one. I wrote out the sum of 30 terms in hopes of finding a pattern. I graphed the first couple positive integer x-coordinates, and found that the sum of the integers above the line is $30+29+28+28+27+26+26 \ldots+ 10$ . The even integer repeats itself every third term in the sum. I found that the average of each of the terms is 20, and there are 30 of them which means that exactly 600 above the line as desired. This give a lower bound because if the slope decreases a little bit, then the points that the line goes through will be above the line.
To find the upper bound, notice that each point with an integer-valued x-coordinate is either $\frac{1}{3}$ or $\frac{2}{3}$ above the line. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates. We turn our attention to $x=28, 29, 30$ which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$ . The point (30,20) is already counted below the line, and we can clearly see that if we slowly increase the slope of the line, we will hit the point (28,19) since (28, $\frac{56}{3}$ ) is closer to the lattice point. The slope of the line which goes through both the origin and (28,19) is $y=\frac{19}{28}x$ . This gives an upper bound of $m=\frac{19}{28}$
Taking the upper bound of m and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ . This is answer $1+84=$ $\boxed{85}$ | E | 85 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | As the procedure shown in the Solution 1, the lower bound of $m$ is $\frac{2}{3}.$ Here I give a more logic way to show how to find the upper bound of $m.$ Denote $N=\sum_{x=1}^{30}(\lfloor mx \rfloor)$ as the number of lattice points in $S$
$N = \lfloor m \rfloor+\lfloor 2m \rfloor+\lfloor 3m \rfloor+\cdots+\lfloor 30m \rfloor = 300 .$
Let $m = \frac{2}{3}+k$ . for $\forall x_{i}\le 30, x\in N^{*}, \lfloor mx_{i} \rfloor = \lfloor \frac{2}{3}x+xk \rfloor.$
Our target is finding the minimum value of $k$ which can increase one unit of $\lfloor mx_{i} \rfloor .$
Notice that:
When $x_{i} = 3n, \lfloor mx_{i} \rfloor = \lfloor 2n+(3n)k \rfloor$ We don't have to discuss this case.
When $x_{i} = 3n+1, \lfloor mx_{i} \rfloor = \lfloor 2n+\frac{2}{3}+(3n+1)k \rfloor, k_{min1}=\frac{1}{3(3n+1)}.$
When $x_{i} = 3n+2, \lfloor mx_{i} \rfloor = \lfloor 2n+1+\frac{1}{3}+(3n+2)k \rfloor, k_{min2}=\frac{2}{3(3n+2)}.$
Here $n\in N^{*}, n \le 9.$
Denote $k_{min}=min\left \{k_{min1},k_{min2} \right \}.$
Obviously $k_{min1}$ and $k_{min2}$ are decreasing. Let's considering the situation when $n=9.$
$k_{min}=min\left\{\frac{1}{84},\frac{2}{87}\right\}=\frac{1}{84}.$
This means that the answer is just $\frac{1}{84}$ , so $a+b=85$ . Choose $\boxed{85}.$ | E | 85 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$ $(p,0)$ $(p,q)$ $(0,q)$ . Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$(p+1)(q+1)$ is the total number of lattice points inside the rectangle. Subtract the number of lattice points on the diagonal, divided by 2 is the number of lattice points below the diagonal, add the number of lattice points on the diagonal, and subtract the lattice points on the $x$ axis, then we get the total number of lattice points on or below the diagonal and $x \ge 1$
There are $900$ lattice points in total. $300$ is $\frac{1}{3}$ of $900$ . The $x$ coordinate of the top-right vertex of the rectangle is $30$ $\frac{1}{2} \cdot 30 \cdot 20 = 300$ . I guess the $y$ coordinate of the top-right vertex of the rectangle is $20$ . Now I am going to verify that. The slope of the diagonal is $\frac{20}{30} = \frac{2}{3}$ , there are $11$ lattice points on the diagonal. Substitute $(p,q)=(30, 20)$ $d=11$ to the above formula:
Because there are $11$ lattice points on line $y = \frac{2}{3}x$ , if $m < \frac{2}{3}$ , then the number of lattice points on or below the line is less than $300$ . So $m = \frac{2}{3}$ is the lower bound.
Now I am going to calculate the upper bound. From $\frac{b}{a} < \frac{b+1}{a+1}$
$\frac{2}{3} = \frac{18}{27} < \frac{19}{28}$
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
[asy] /* Created by Brendanb4321, modified by isabelchen */ import graph; size(18cm); defaultpen(fontsize(9pt)); xaxis(0,31,Ticks(1.0)); yaxis(0,22,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30*19/28), dotted); draw((0,0)--(31,31*21/31), dotted); for (int i = 1; i<=31; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); dot((31,21), blue); label("$m=2/3$", (33,20)); label("$m=21/31$", (33,21)); label("$m=19/28$", (33,22)); [/asy]
If $m = \frac{21}{31}$ , I will calculate by using the rectangle with blue vertex $(p,q) = (31, 21)$ , then subtract lattice points on line $x = 31$ , which is $21$ . There are 2 lattice points on the diagonal, $d=2$
If $m = \frac{19}{28}$ , I will calculate by using the rectangle with red vertex $(p,q) = (28, 19)$ , then add lattice points on line $x = 29$ and $x = 30$ , which is $19 + 20 = 39$ . There are 2 lattice points on the diagonal, $d=2$
When $m$ increases, more lattice points falls below the line $y = mx$ . Any value larger than $\frac{19}{28}$ has more than $301$ lattice points on or below $y = \frac{19}{28} x$ . So the upper bound is $\frac{19}{28}$
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ $\boxed{85}$ | E | 85 |
633c7a578d996d6c6a02a3f0457eaac2 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | The lower bound of $m$ is $\frac23 = \frac{20}{30}$ . Inside the rectangle with vertices $(0,0)$ $(30,0)$ $(30,20)$ $(0, 20)$ and diagonal $y = \frac23 x$ , there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551 + 9 = 560$ . Half of the $560$ lattice points are below diagonal $y = \frac23 x$ $560 \cdot \frac12 = 280$ . There are $20$ lattice points on edge $x = 30$ $280 + 20 = 300$ . Once $m < \frac23$ , the $9$ lattice points on diagonal $y = \frac23 x$ will be above the new diagonal, making the number of lattice points on and below the diagonal less than $300$
Now we are going to calculate the upper bound by the following formula:
[asy] /* Created by isabelchen */ import graph; size(8cm); defaultpen(fontsize(9pt)); xaxis(0,8); yaxis(0,6); draw((0,0)--(7,5)); draw((7,0)--(7,5)); draw((0,5)--(7,5)); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6)); label("$(0,0)$", (0,0), SW); label("$(a, b)$", (7,5), NE); dot((7,0)); label("$a$", (7,0), S); dot((0,5)); label("$b$", (0,5), W); [/asy]
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
When $a = 31$ $b = 21$ $\frac{(31-1)(21-1)}{2} = 300$
When $a = 28$ $b = 19$ $\frac{(28-1)(19-1)}{2} = 243$ . Below the line $y = \frac{19}{28} x$ , there are $19$ lattice points on line $x = 28$ $19$ lattice points on line $x = 29$ $20$ lattice points on line $x = 30$ $243 + 19 + 19 + 20 = 301$
More lattice points fall below the line $y = mx$ as $m$ increases. There are more than $301$ lattice points on and below the line for any $m$ greater than $\frac{19}{28}$ . Therefore, the upper bound is $\frac{19}{28}$
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ , so $1+84=\boxed{85}$ | E | 85 |
2a070373c59e4bfe58a7696b2c94dfd6 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_2 | The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$ . What is the average of $a$ and $b$
$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$ | The arithmetic mean of the numbers $3, 5, 7, a,$ and $b$ is equal to $\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15$ . Solving for $a+b$ , we get $a+b=60$ . Dividing by $2$ to find the average of the two numbers $a$ and $b$ gives $\frac{60}{2}=\boxed{30}$ | C | 30 |
8ac11081c69023fb62de5acf84ec676b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_3 | Assuming $a\neq3$ $b\neq4$ , and $c\neq5$ , what is the value in simplest form of the following expression? \[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}$ | If $x\neq y,$ then $\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression: \[\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c-5}}}{\color{green}\underset{1}{\cancel{4-b}}}=(-1)(-1)(-1)=\boxed{1}.\] ~CoolJupiter ~MRENTHUSIASM | A | 1 |
8ac11081c69023fb62de5acf84ec676b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_3 | Assuming $a\neq3$ $b\neq4$ , and $c\neq5$ , what is the value in simplest form of the following expression? \[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}$ | At $(a,b,c)=(4,5,6),$ the answer choices become
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } {-}\frac{1}{120} \qquad \textbf{(E) } \frac{1}{120}$
and the original expression becomes \[\frac{-1}{1}\cdot\frac{-1}{1}\cdot\frac{-1}{1}=\boxed{1}.\] ~MRENTHUSIASM | A | 1 |
0abc4ab743b94eebaf4c89a5e03ba9ed | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_4 | A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26$ | Since the driver travels $60$ miles per hour and each hour she uses $2$ gallons of gasoline, she spends $$4$ per hour on gas. If she gets $$0.50$ per mile, then she gets $$30$ per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $\boxed{26}$ .
~mathsmiley | E | 26 |
0abc4ab743b94eebaf4c89a5e03ba9ed | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_4 | A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26$ | The driver is driving for $2$ hours at $60$ miles per hour, she drives $120$ miles. Therefore, she uses $\frac{120}{30}=4$ gallons of gasoline. So, total she has $$0.50\cdot120-$2.00\cdot4=$60-$8=$52$ . So, her rate is $\frac{52}{2}=\boxed{26}$ .
~sosiaops | E | 26 |
b40e02d22d26944e2dfe4334c7a3569c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 | What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$ | Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
Case 1:
The equation yields $x^2-12x+34=2$ , which is equal to $(x-4)(x-8)=0$ . Therefore, the two values for the positive case is $4$ and $8$
Case 2:
Similarly, taking the nonpositive case for the value inside the absolute value notation yields $x^2-12x+34=-2$ . Factoring and simplifying gives $(x-6)^2=0$ , so the only value for this case is $6$
Summing all the values results in $4+8+6=\boxed{18}$ | C | 18 |
b40e02d22d26944e2dfe4334c7a3569c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 | What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$ | We have the equations $x^2-12x+32=0$ and $x^2-12x+36=0$
Notice that the second is a perfect square with a double root at $x=6$ , and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is $-(-12)$ or $12$ $12+6=\boxed{18}$ | C | 18 |
b40e02d22d26944e2dfe4334c7a3569c | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 | What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$ | Completing the square gives \begin{align*} \left|(x-6)^2-2\right|&=2 \\ (x-6)^2-2&=\pm2. \hspace{15mm}(\bigstar) \end{align*} Note that the graph of $y=(x-6)^2-2$ is an upward parabola with the vertex $(6,-2)$ and the axis of symmetry $x=6;$ the graphs of $y=\pm2$ are horizontal lines.
We apply casework to $(\bigstar):$
Finally, the sum of all solutions is $12+6=\boxed{18}.$ | C | 18 |
12029877d9f3f90d24f177f850e5b4de | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_6 | How many $4$ -digit positive integers (that is, integers between $1000$ and $9999$ , inclusive) having only even digits are divisible by $5?$
$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$ | The units digit, for all numbers divisible by 5, must be either $0$ or $5$ . However, since all digits are even, the units digit must be $0$ . The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices for each of the middle 2 digits, and 4 choices for the thousands digit, we have a total of $4\cdot5\cdot5\cdot1 = \boxed{100} \qquad$ numbers. | B | 100 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\frac{50}{5}=\boxed{10}$ | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \boxed{10}$ as our answer.
~Baolan | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\cdot2$ , or $\boxed{10}$ .
~Arctic_Bunny, edited by KINGLOGIC | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Let us consider the horizontal rows. Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ . Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\boxed{10}$ .
~cw357 | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | The mean of the set of numbers is $(14-10) \div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)
One row of the square would be \[\square \square 2 \square \square\]
Adding the numbers would be
\[0, 1, 2, 3, 4\]
with a sum of $\boxed{10}$ | C | 10 |
f7fce36a24659f563b87f5196ca5722b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_7 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\boxed{10}$ .
~MC413551 | null | 10 |
bc958ce2333164b4e62bdad7adc0b369 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | Looking at the numbers, you see that every set of $4$ has $3$ positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us $2+10+18...+394$ . Clearly, this pattern is an arithmetic sequence. By using the formula we get $\frac{2+394}{2}\cdot 50=\boxed{9900}$ | B | 9900 |
bc958ce2333164b4e62bdad7adc0b369 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of $50\cdot (-2)=-100$ . Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to $100^2=10000$ . Adding these two, we obtain the answer of $\boxed{9900}$ | B | 9900 |
bc958ce2333164b4e62bdad7adc0b369 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_8 | What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$ | We can break this entire sum down into $4$ integer bits, in which the sum is $2x$ , where $x$ is the first integer in this bit. We can find that the first sum of every sequence is $4x-3$ , which we plug in for the $50$ bits in the entire sequence is $1+2+3+\cdots+50=1275$ , so then we can plug it into the first term of every sequence equation we got above $4(1275)-3(50)=4950$ , and so the sum of every bit is $2x$ , and we only found the value of $x$ , the sum of the sequence is $4950\cdot2=\boxed{9900}$ .-middletonkids | B | 9900 |
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