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cb010b64769a59dd885495b2be9d59f6 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_24 | Let $f_0(x)=x+|x-100|-|x+100|$ , and for $n\geq 1$ , let $f_n(x)=|f_{n-1}(x)|-1$ . For how many values of $x$ is $f_{100}(x)=0$
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | First, notice that the recursive rule moves the current value $1$ closer to $0$ . Upon reaching $0$ , it alternates between $-1$ and $0$ . This means that $f_{100}(x) = 0$ exactly when $|f_0(x)| \le 100$ (to reach $0$ in time) and $f_0(x)$ is even (so $f_{100}(x) \ne -1$ ).
Casework each part of $f_0(x)$ (where the expressions in the absolute values do not change sign): \[x \le -100 \implies f_0(x) = x-(x-100)+(x+100) = x+200\] so even $-300 \le x \le -100$ work. \[-100 \le x \le 100 \implies f_0(x) = x-(x-100)-(x+100) = -x\] so even $-100 \le x \le 100$ work. \[100 \le x \implies f_0(x) = x+(x-100)-(x+100) = x-200\] so even $100 \le x \le 300$ work.
Putting these together, all even $x$ where $-300 \le x \le 300$ work. So, the answer is $2\cdot150+1 = \boxed{301}$ .
~revised by emerald_block | C | 301 |
6e12f8774789a435e63bd4b414439f22 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_25 | The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$ . For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$
$\textbf{(A) }38\qquad \textbf{(B) }40\qquad \textbf{(C) }42\qquad \textbf{(D) }44\qquad \textbf{(E) }46\qquad$ | The axis of $P$ is inclined at an angle $\theta$ relative to the coordinate axis, where $\tan\theta = \tfrac 34$ . We rotate the coordinate axis by angle $\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\widetilde{x}, \widetilde{y})$ be the coordinates in the rotated system. Then $(x,y)$ and $(\widetilde{x}, \widetilde{y})$ are related by \begin{align} \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\ y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y} \end{align} In the rotated coordinate system, the parabola has focus at $(0,0)$ and the two points on it are at $(5,0)$ and $(-5,0)$ . Therefore, the directrix is $\widetilde{y}=\pm 5$ ; we can, WLOG, choose $\widetilde{y}=-5$ . For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is \begin{align}\tag{2} \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25) \end{align} From $(1)$ we have $|4x+3y|=5\widetilde{x}$ , so we need $|\widetilde{x}|<200$ . Substituting $(2)$ in $(1)$ , we get \begin{align*} 50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\ 50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100 \end{align*} For $x$ to be an integer $\widetilde{x}$ must be a multiple of 5; setting $\widetilde{x}=5a$ we get \[2x = 8a - 3 a^2 + 3\] Now we need $a$ to be odd, i.e. $\widetilde{x}=5a$ is an odd multiple of $5$ , in which case we get $y = 3 a + 2 a^2 - 2$ , which is also an integer. The values that satisfy the given conditions correspond to $\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}$ , and there are $\boxed{40}$ such numbers. | B | 40 |
b5487dbfbd2d7bb6ed7adc7df24a58c9 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_1 | Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 39\qquad\textbf{(E)}\ 41$ | She has $p$ pennies and $n$ nickels, where $n + p = 13$ . If she had $n+1$ nickels then $n+1 = p$ , so $2n+ 1 = 13$ and $n=6$ . So she has 6 nickels and 7 pennies, which clearly have a value of $\boxed{37}$ cents. | C | 37 |
4fbc611b5b09b189e9d5144f6a0eb995 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_2 | Orvin went to the store with just enough money to buy $30$ balloons. When he arrived he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 38\qquad\textbf{(E)}\ 39$ | If every balloon costs $n$ dollars, then Orvin has $30n$ dollars. For every balloon he buys for $n$ dollars, he can buy another for $\frac{2n}{3}$ dollars. This means it costs him $\frac{5n}{3}$ dollars to buy a bundle of $2$ balloons. With $30n$ dollars, he can buy $\frac{30n}{\frac{5n}{3}} = 18$ sets of two balloons, so the total number of balloons he can buy is $18\times2 \implies \boxed{36}$ | C | 36 |
82c160e21190d12cb2c053c43b0a7bca | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_5 | Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$ , and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill((6,0)--(8,0)--(8,26)--(6,26)--cycle,gray); fill((12,0)--(14,0)--(14,26)--(12,26)--cycle,gray); fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray); fill((24,0)--(26,0)--(26,26)--(24,26)--cycle,gray); fill((0,0)--(26,0)--(26,2)--(0,2)--cycle,gray); fill((0,12)--(26,12)--(26,14)--(0,14)--cycle,gray); fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); [/asy]
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$ | Let the height of the panes equal $5x$ , and let the width of the panes equal $2x$ . Now notice that the total width of the borders equals $10$ , and the total height of the borders is $6$ . We have \[10 + 4(2x) = 6 + 2(5x)\] \[x = 2\] Now, the total side length of the window equals \[10+ 4(2x) = 10 + 16 = \boxed{26}\] | A | 26 |
08e34686f2ba7f1edffbb15f89620549 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_6 | Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume $\frac{3}{4}$ of their drinks, Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both drank the same amount. How many ounces of lemonade did they drink together?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$ | Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\frac{3}{2}x$ ounces. After both consume $\frac{3}{4}$ of their drinks, Ed and Ann have $\frac{x}{4}$ and $\frac{3x}{8}$ ounces of their drinks remaining. Ann gives away $\frac{x}{8} + 2$ ounces to Ed.
In the end, Ed drank everything in his original lemonade plus what Ann gave him, and Ann drank everything in her original lemonade minus what she gave Ed. Thus we have \[x + \frac{x}{8} + 2 = \frac{3x}{2} - \frac{x}{8} - 2\] \[x = 16\] The total amount the two of them drank is simply \[x + \frac{3}{2}x = 16 + 24 = \boxed{40}\] | D | 40 |
500b614dfd1dcebc41636280d48621ac | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_7 | For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | We know that $n \le 30$ or else $30-n$ will be negative, resulting in a negative fraction. We also know that $n \ge 15$ or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values $n$ from $15$ to $30$ gives us integer values for $n=15, 20, 24, 25, 27, 28, 29$ . Counting them up, we have $\boxed{7}$ possible values for $n$ | D | 7 |
500b614dfd1dcebc41636280d48621ac | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_7 | For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let $\frac{n}{30-n}=m$ , where $m \in \mathbb{N}$ . Solving for $n$ , we find that $n=\frac{30m}{m+1}$ . Because $m$ and $m+1$ are relatively prime, $m+1|30$ . Our answer is the number of proper divisors of $2^13^15^1$ , which is $(1+1)(1+1)(1+1)-1 = \boxed{7}$ | D | 7 |
f5e2e8135ba1335fb59d277ebadb292f | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_8 | In the addition shown below $A$ $B$ $C$ , and $D$ are distinct digits. How many different values are possible for $D$
\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$ . We know that $A+B = D$ . Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two distinct natural numbers. Letting $A=1$ , and letting $B=2,3,4,5,6,7,8$ , we have \[D = 3,4,5,6,7,8,9 \implies \boxed{7}\] | C | 7 |
8547f5e8bbc32e4d9fba8ac6b9686c2e | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_9 | Convex quadrilateral $ABCD$ has $AB=3$ $BC=4$ $CD=13$ $AD=12$ , and $\angle ABC=90^{\circ}$ , as shown. What is the area of the quadrilateral?
[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,C,25)); [/asy]
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5$ | Note that by the pythagorean theorem, $AC=5$ . Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to \[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{36}\] | B | 36 |
e469a07a09e815d0fbeda3630fcc9cd1 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_10 | Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41$ | We know that the number of miles she drove is divisible by $5$ , so $a$ and $c$ must either be the equal or differ by $5$ . We can quickly conclude that the former is impossible, so $a$ and $c$ must be $5$ apart. Because we know that $c > a$ and $a + c \le 7$ and $a \ge 1$ , we find that the only possible values for $a$ and $c$ are $1$ and $6$ , respectively. Because $a + b + c \le 7$ $b = 0$ . Therefore, we have \[a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{37}\] | D | 37 |
e469a07a09e815d0fbeda3630fcc9cd1 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_10 | Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41$ | Let the number of hours Danica drove be $k$ . Then we know that $100a + 10b + c + 55k$ $100c + 10b + a$ . Simplifying, we have $99c - 99a = 55k$ , or $9c - 9a = 5k$ . Thus, k is divisible by $9$ . Because $55 * 18 = 990$ $k$ must be $9$ , and therefore $c - a = 5$ . Because $a + b + c \leq{7}$ and $a \geq{1}$ $a = 1$ $c = 6$ and $b = 0$ , and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$ , or $\boxed{37}$ | D | 37 |
366201ecdaa2b38851615ec29fd89818 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_11 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$ | We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$ . In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$ . If we let the list be $1,2,3,8,8,9,10,11,12,13,j$ , then $j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$
Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$ . Following the same process as above, we get $j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$ . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{35}$ | E | 35 |
14eb9cc0ce61a1783d6a5d1a3594ca03 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_12 | A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \ge b \ge c$ $b+c > a$ , and $a, b, c < 5$ . Now we enumerate the elements of $T$
$(4, 4, 4)$
$(4, 4, 3)$
$(4, 4, 2)$
$(4, 4, 1)$
$(4, 3, 3)$
$(4, 3, 2)$
$(3, 3, 3)$
$(3, 3, 2)$
$(3, 3, 1)$
$(3, 2, 2)$
$(2, 2, 2)$
$(2, 2, 1)$
$(1, 1, 1)$
It should be clear that $|S|$ is simply $|T|$ minus the larger "duplicates" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$ ). Since $|T|$ is $13$ and the number of higher duplicates is $4$ , the answer is $13 - 4$ or $\boxed{9}$ | B | 9 |
3cf80ad9fb74e6affdb5cbf43477ac5c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_16 | Let $P$ be a cubic polynomial with $P(0) = k$ $P(1) = 2k$ , and $P(-1) = 3k$ . What is $P(2) + P(-2)$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k$ | Let $P(x) = Ax^3+Bx^2 + Cx+D$ . Plugging in $0$ for $x$ , we find $D=k$ , and plugging in $1$ and $-1$ for $x$ , we obtain the following equations: \[A+B+C+k=2k\] \[-A+B-C+k=3k\] Adding these two equations together, we get \[2B=3k\] If we plug in $2$ and $-2$ in for $x$ , we find that \[P(2)+P(-2) = 8A+4B+2C+k+(-8A+4B-2C+k)=8B+2k\] Multiplying the third equation by $4$ and adding $2k$ gives us our desired result, so \[P(2)+P(-2)=12k+2k=\boxed{14}\] | E | 14 |
9cfe3f60cd374c069df86eeb8babeea7 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$ | Let $y = m(x - 20) + 14$ . Equating them:
$x^2 = mx - 20m + 14$
$x^2 - mx + 20m - 14 = 0$
For there to be no solutions, the discriminant must be less than zero:
$m^2 - 4(20m - 14) < 0$
$m^2 - 80m + 56 < 0$
So $m < 0$ for $r < m < s$ where $r$ and $s$ are the roots of $m^2 - 80m + 56 = 0$ and their sum by Vieta's formulas is $\boxed{80}$ | E | 80 |
9cfe3f60cd374c069df86eeb8babeea7 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$ | The line will begin to intercept the parabola when its slope equals that of the parabola at the point of tangency. Taking the derivative of the equation of the parabola, we get that the slope equals $2x$ . Using the slope formula, we find that the slope of the tangent line to the parabola also equals $\frac{14-x^2}{20-x}$ . Setting these two equal to each other, we get \[2x = \frac{14-x^2}{20-x} \implies x^2-40x+14 = 0\] Solving for $x$ , we get \[x= 20\pm \sqrt{386}\] The sum of the two possible values for $x$ where the line is tangent to the parabola is $40$ , and the sum of the slopes of these two tangent lines is equal to $2x$ , or $\boxed{80}$ | E | 80 |
9cfe3f60cd374c069df86eeb8babeea7 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$ | The smaller solution is basically negligible in comparison with the solution with the larger slope. Try some values like of $x^2 = y$ where $x = 40$ => $y = 1600$ , and slope ~80. Trying a few values leads us to conclude the least possible value is around $80$ , so the answer is $\boxed{80}$ | E | 80 |
77e0fabb4f8f99b4274e980292ddff5d | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_24 | problem_id
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
Name: Text, dtype: object | We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$ . By choosing the full circle, we can obtain $15$ . By choosing everything except for $1, 2, 3, 4,$ and $5$ , we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$
This means that we now only need to check for $6, 7, 8,$ and $9$ . However, once we have found a set summing to $6$ , we can choose all remaining numbers and obtain a set summing to $15-6=9$ , and similarly for $7$ and $8$ . Thus, we only need to check each case for whether or not we can obtain $6$ or $7$
We can make $6$ by having $4, 2$ , or $3, 2, 1$ , or $5, 1$ . We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$x$", A, N); label("$y$", C, SW); label("$z$", D, SE); [/asy]
Now, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$ , because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$2$", C, SW); label("$1$", D, SE); [/asy]
We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$ , so we must place $4$ next to the $1$ and $5$ next to the $2$
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$5$", B, NW); label("$2$", C, SW); label("$1$", D, SE); label("$4$", E, NE); [/asy]
This is the only solution to make $6$ "bad."
Next we move on to $7$ , which can be made by $3, 4$ , or $5, 2$ , or $4, 2, 1$ . We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.
[asy] draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$1$", A, N); label("$3$", B, NW); label("$2$", C, SW); label("$4$", D, SE); label("$5$", E, NE); [/asy]
This is the only solution to make $7$ "bad."
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\boxed{2}$ | B | 2 |
c298325224d0e2c8a3d0be9dd4c791a8 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_20 | For how many positive integers $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$
$\textbf{(A) }10\qquad \textbf{(B) }18\qquad \textbf{(C) }19\qquad \textbf{(D) }20\qquad \textbf{(E) }\text{infinitely many}\qquad$ | The domain of the LHS implies that \[40<x<60\] Begin from the left hand side \[\log_{10}[(x-40)(60-x)]<2\] \[(x-40)(60-x)<100\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{18}$ integers. | B | 18 |
de4ed88020c057f81c6cc0cfae94254c | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_23 | The number $2017$ is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$ . What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$ | Note that $2014\equiv -3 \mod2017$ . We have for $k\ge1$ \[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\] \[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\] Therefore \[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\] This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{ if n is odd}\] and \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{ if n is even}\] Obviously, $62$ falls in the second category, so our desired value is \[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{1024}\] | C | 1024 |
f0271e063089f1f25ec5034e05e9b35a | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_25 | Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$ | Rewrite $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ . Now let $a = \cos{2x}$ , and let $b = \cos{\left( \frac{2014\pi^2}{x} \right) }$ . We have: \[2a(a - b) = 2a^2 - 2\]
Therefore, \[ab = 1\]
Notice that either $a = 1$ and $b = 1$ or $a = -1$ and $b = -1$ . For the first case, $a = 1$ only when $x = k\pi$ and $k$ is an integer. $b = 1$ when $\frac{2014\pi^2}{k\pi}$ is an even multiple of $\pi$ , and since $2014 = 2*19*53$ $b =1$ only when $k$ is an odd divisor of $2014$ . This gives us these possible values for $x$ \[x= \pi, 19\pi, 53\pi, 1007\pi\] For the case where $a = -1$ $\cos{2x} = -1$ , so $x = \frac{m\pi}{2}$ , where m is odd. $\frac{2014\pi^2}{\frac{m\pi}{2}}$ must also be an odd multiple of $\pi$ in order for $b$ to equal $-1$ , so $\frac{4028}{m}$ must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for $m$ , and therefore no cases where $a = -1$ and $b = -1$ . Therefore, the sum of all our possible values for $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{1080}\] | D | 1080 |
f0271e063089f1f25ec5034e05e9b35a | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_25 | Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$ | Rewriting $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ and transposing $2\cos{2x}$ from the LHS to the RHS, we get,
\[\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}\] \[\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RHS}}\]
By the AM-GM Inequality
\[\cos{2x} + \frac{1}{\cos{2x}} \in (-\infty, -2] \cup [2, \infty)\]
Also, because of the range of $\cos$
\[1 + \cos{\left(\frac{2014\pi^2}{x}\right)} \in [0, 2]\]
Hence, $\text{LHS} = \text{RHS} = 2$ , and we get ( $m, n \in \mathbb{Z}$ ),
From $(1)$ and $(2)$ \[m = \frac{1007}{n}\] \[\implies m \in \{1, 19, 53, 1007\}\] \[\implies x \in \{1\pi, 19\pi, 53\pi, 1007\pi\}\]
Therefore, sum of values of $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{1080}\] | D | 1080 |
f7fe5848711c7b7cad5dd6540551e1f3 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_5 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ $105$ , Dorothy paid $ $125$ , and Sammy paid $ $175$ . In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: \[105 + t = 125 + d.\] Rearranging, we have \[t-d = \boxed{20}.\] | B | 20 |
f7fe5848711c7b7cad5dd6540551e1f3 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_5 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ $105$ , Dorothy paid $ $125$ , and Sammy paid $ $175$ . In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | Add up the amounts that Tom, Dorothy, and Sammy paid to get $ $405$ , and divide by 3 to get $ $135$ , the amount that each should have paid.
Tom, having paid $ $105$ , owes Sammy $ $30$ , and Dorothy, having paid $ $125$ , owes Sammy $ $10$
Thus, $t - d = 30 - 10 = 20$ , which is $\boxed{20}$ | B | 20 |
221460b070007279e52d8c55743ed3ee | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_6 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$ | (similar to Solution 1, however a slightly more obvious way)
Say that
x = # of 2-pt shots
y = # of 3-pt shots
Because the total number of shots is $30$ $x + y = 30$
However, Shenille was only successful on $20\%$ of the 3-pt shots, and $30\%$ of the 2-pt shots, so
$0.2x + 0.3y$ = #number of successful shots
For each successful shot, there is an associated number of points with it.
Therefore, $0.2(x)(3) + (0.3)(y)(2)$ = her score
this evaluates to $0.6 (x + y)$ = her score
$x + y$ is already determined to be 30, so her score is $0.6 (30) = \boxed{18}$ | B | 18 |
09eeabd884c1215a1eef77480102c030 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_7 | The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, \[S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.\] Suppose that $S_9 = 110$ and $S_7 = 42$ . What is $S_4$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad$ | $S_9 = 110$ $S_7 = 42$
$S_8 = S_9 - S_ 7 = 110 - 42 = 68$
$S_6 = S_8 - S_7 = 68 - 42 = 26$
$S_5 = S_7 - S_6 = 42 - 26 = 16$
$S_4 = S_6 - S_5 = 26 - 16 = 10$
Therefore, the answer is $\boxed{10}$ | C | 10 |
9a3294873b76f92e707d11dd5f9f787a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | $x+\tfrac{2}{x}= y+\tfrac{2}{y}$
Since $x\not=y$ , we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$
Cross multiply in either equation, giving us $xy=2$
$\boxed{2}$ | D | 2 |
9a3294873b76f92e707d11dd5f9f787a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | Let $A = x + \frac{2}{x} = y + \frac{2}{y}.$ Consider the equation \[u + \frac{2}{u} = A.\] Reorganizing, we see that $u$ satisfies \[u^2 - Au + 2 = 0.\] Notice that there can be at most two distinct values of $u$ which satisfy this equation, and $x$ and $y$ are two distinct possible values for $u.$ Therefore, $x$ and $y$ are roots of this quadratic, and by Vieta’s formulas we see that $xy$ thereby must equal $\boxed{2}.$ | null | 2 |
9a3294873b76f92e707d11dd5f9f787a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | \[x + \frac{2}{x} = y + \frac{2}{y}.\]
Multiply both sides by xy to get
\[x^2y + 2y = y^2x +2x\]
Rearrange to get
\[x^2y - y^2x = 2x - 2y\]
Factor out $xy$ on the left side and $2$ on the right side to get
\[xy(x-y) = 2(x-y)\]
Divide by $(x-y)$ {You can do this since x and y are distinct} to get
$\boxed{2}$ ~ e__ (the goat) | D | 2 |
503cf642d857d9a7f0d4b88cbc2a7ca6 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_9 | In $\triangle ABC$ $AB=AC=28$ and $BC=20$ . Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively. What is the perimeter of parallelogram $ADEF$
[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]
$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$ | Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $\triangle ABC$ , the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$ , and are therefore also isosceles triangles.
It follows that $BD = DE$ . Thus, $AD + DE = AD + DB = AB = 28$
Since opposite sides of parallelograms are equal, the perimeter is $2 * (AD + DE) = \boxed{56}$ | C | 56 |
503cf642d857d9a7f0d4b88cbc2a7ca6 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_9 | In $\triangle ABC$ $AB=AC=28$ and $BC=20$ . Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively. What is the perimeter of parallelogram $ADEF$
[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]
$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$ | We can set point $F$ to be on point $C$ , and point $D$ to be on point $A$
This makes a degenerate parallelogram with sides of length $28$ and $0$ , so it has a perimeter of $28 + 28 = \boxed{56}$ | C | 56 |
39dd8bce5448ecf372ef6cf13a7e12bd | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_10 | Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$ | Let us begin by working with the condition $0.\overline{ab} = 0.ababab\cdots,$ . Let $x = 0.ababab\cdots$ . So, $100x-x = ab \Rightarrow x = \frac{ab}{99}$ . In order for this fraction $x$ to be in the form $\frac{1}{n}$ $99$ must be a multiple of $ab$ . Hence the possibilities of $ab$ are $1,3,9,11,33,99$ . Checking each of these, $\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},$ and $\frac{99}{99} = 1$ . So the only values of $n$ that have distinct $a$ and $b$ are $11,33,$ and $99$ . So, $11+33+99= \boxed{143}$ | D | 143 |
39dd8bce5448ecf372ef6cf13a7e12bd | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_10 | Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$ | Notice that we have $\frac{100}{n}= ab.\overline{ab}$
We can subtract $\frac{1}{n}=00.\overline{ab}$ to get \[\frac{99}{n}=ab\]
From this we determine $n$ must be a positive factor of $99$
The factors of $99$ are $1,3,9,11,33,$ and $99$
For $n=1,3,$ and $9$ however, they yield $ab=99,33$ and $11$ which doesn't satisfy $a$ and $b$ being distinct.
For $n=11,33$ and $99$ we have $ab=09,03$ and $01$ . (Notice that $a$ or $b$ can be zero)
The sum of these $n$ are $11+33+99=143$
$\boxed{143}$ | D | 143 |
753f911cd6cd70358c3904fdad2ec650 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_13 | Let points $A = (0,0) , \ B = (1,2), \ C = (3,3),$ and $D = (4,0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left (\frac{p}{q}, \frac{r}{s} \right )$ , where these fractions are in lowest terms. What is $p + q + r + s$
$\textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75$ | If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.
Pick's Theorem states that
$A$ $I$ $+$ $\frac{B}{2}$ $1$ , where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice points on the boundary of the polygon.
In this case,
$A$ $5$ $+$ $\frac{7}{2}$ $1$ $7.5$
so
$\frac{A}{2}$ $3.75$
The bottom half of the quadrilateral makes a triangle with base $4$ and half the total area, so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for its area to be $3.75$ . This height is the y coordinate of our desired intersection point.
Note that segment CD lies on the line $y = -3x + 12$ . Substituting in $\frac{15}{8}$ for y, we can find that the x coordinate of our intersection point is $\frac{27}{8}$
Therefore the point of intersection is ( $\frac{27}{8}$ $\frac{15}{8}$ ), and our desired result is $27+8+15+8= \boxed{58}$ | B | 58 |
753f911cd6cd70358c3904fdad2ec650 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_13 | Let points $A = (0,0) , \ B = (1,2), \ C = (3,3),$ and $D = (4,0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left (\frac{p}{q}, \frac{r}{s} \right )$ , where these fractions are in lowest terms. What is $p + q + r + s$
$\textbf{(A)} \ 54 \qquad \textbf{(B)} \ 58 \qquad \textbf{(C)} \ 62 \qquad \textbf{(D)} \ 70 \qquad \textbf{(E)} \ 75$ | Let the point of intersection be $E$ , with coordinates $(x, y)$ . Then, $ABCD$ is cut into $ABCE$ and $AED$
Since the areas are equal, we can use Shoelace Theorem to find the area. This gives $3 + 3x - 3y = 4y$
The line going through $CD$ is $y = -3x + 12$ . Since $E$ is on $CD$ , we can substitute this in, giving $3 + 3x = -21x + 84$ . Solving for $x$ gives $\frac{27}{8}$ . Plugging this back into the line equation gives $y = \frac{15}{8}$ , for a final answer of $\boxed{58}$ | null | 58 |
a45de7580445f2009a57ff5d32c09b07 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_15 | Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
$\textbf{(A)} \ 96 \qquad \textbf{(B)} \ 108 \qquad \textbf{(C)} \ 156 \qquad \textbf{(D)} \ 204 \qquad \textbf{(E)} \ 372$ | We tackle the problem by sorting it by how many stores are involved in the transaction.
1) 2 stores are involved. There are $\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \cdot 2 = 12$ total arrangements.
2) 3 stores are involved. There are $\binom{4}{3} = 4$ ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately.
Separately: All children must be in one store. There are $3!$ ways to arrange this. $6$ ways in total.
Together: Both parents are in one store and the 3 children are split between the other two. There are $\binom{3}{2}$ ways to split the children and $3!$ ways to choose to which store each group will be sold. $3! \cdot \binom{3}{2} = 18$
$(6 + 18) \cdot 4 = 96$ total arrangements.
3) All 4 stores are involved. We break down the problem as previously shown.
Separately: All children must be split between two stores. There are $\binom{3}{2} = 3$ ways to arrange this. We can then arrange which group is sold to which store in $4!$ ways. $4! \cdot 3 = 72$
Together: Both parents are in one store and the 3 children are each in another store. There are $4! = 24$ ways to arrange this.
$24 + 72 = 96$ total arrangements.
Final Answer: $12 + 96 + 96 = \boxed{204}$ | D | 204 |
03b042f010cad61f3c89575a72dd8706 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_16 | $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$ | Let the number of rocks in $A$ be $a$ $B$ be $b$ $C$ be $c$ . The total weight of $A$ be $40a$ $B$ be $50b$ $C$ be $kc$
We can write the information given as, $\frac{40a + 50b}{a+b} = 43$ $\frac{40a + kc}{a+c} = 44$ $\frac{50b + kc}{b+c} = ?$
$40a + 50b = 43 a + 43 b$ $3a = 7b$
$40a + kc = 44a + 44 c$ $kc = 4a + 44c = \frac{28}{3}b + 44c$
$\frac{50b + kc}{b+c} = \frac{50 \cdot \frac{3}{7}a + kc}{\frac{3}{7}a+c} = \frac{150a + 7kc}{3a + 7c} = \frac{150a + 28a + 308c}{3a+7c} = \frac{178a + 308c}{3a+7c} = \frac{44(3a+7c)+46a}{3a+7c}$
$= 44 + \frac{46a}{3a+7c} = 44 + \frac{46}{3 + \frac{7}{a}} < 44 + \frac{46}{3} \approx 59.3$
$\frac{50b + kc}{b+c} \le \boxed{59}$ | E | 59 |
03b042f010cad61f3c89575a72dd8706 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_16 | $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$ | Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: \[\frac{40A_n+50B_n}{A_n+B_n}=43\] .
Cross-multiplying, we get: \[40A_n+50B_n=43A_n+43B_n \implies 3A_n=7B_n\] .
Let's say we have $7k$ rocks in $A$ and $3k$ rocks in $B$ . Hence, we have $280k$ and $150k$ as the total weight of piles $A$ and $B$ , respectively. Let the total weight of $C$ be $m$ , and the total number of rocks in $C$ be $n$ .
Using the last restriction regarding the average of piles $A$ and $C$ , we have: \[\frac{280k+m}{7k+n}=44 \implies 280k + m=280k + 28k + 44n \implies m=28k+44n\] .
To find the average of piles $B$ and $C$ , we can establish the expression: \[\frac{150k+28k+44n}{3k+n}=\frac{178k+44n}{3k+n}=\frac{132k+44n+46k}{3k+n}=\frac{44(3k+n)+46k}{3k+n}=44+\frac{46k}{3k+n}.\] When we let the final expression equal $59$ , we get: \[44+\frac{46k}{3k+n}=59\implies\frac{46k}{3k+n}=15\] Cross-multiplying, we get: \[46k=45k+14n\implies k=14n\] $k$ is still positive here, so 59 works. As this is the greatest option, we can circle $\textbf{(E)}$ immediately.
To show why $59$ is the greatest, consider the following:
When we let the final expression equal $60$ , we get: \[44+\frac{46k}{3k+n}=60\implies\frac{46k}{3k+n}=16\] Cross-multiplying, we get: \[46k=48k+14n\implies k=-7n\] Since $k$ is positive, the final expression could not equal 60. It further implies that the final expression could not equal any other integer greater than 60. Therefore, we have our final answer $\boxed{59}$ | null | 59 |
7b93826b436587f2b9c6024c63afbc28 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850$ | The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$
The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}\cdot \frac{11}{12}*x$
Note that
$12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}$
$11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$
All the $2$ s and $3$ s cancel out of $11!$ , leaving
$11 \cdot 5 \cdot 7 \cdot 5 = 1925$
in the numerator.
We know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $\boxed{1925}$ coins for the twelfth pirate. | D | 1925 |
7b93826b436587f2b9c6024c63afbc28 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850$ | The answer cannot be an even number. Here is why:
Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some $\frac{n}{12}$ . This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by $\frac{2}{3}$ for pirate 4, but it immediately drops again in the next step).
Therefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.
Only one of the choices given is odd, $\boxed{1925}$ | D | 1925 |
7b93826b436587f2b9c6024c63afbc28 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850$ | Let $x$ be the number of coins the $12$ th pirate takes. Then the number of coins the $k<12$ th pirate takes is $\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x$ . For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors, obviously there are sufficient $2$ s and $3$ s, so we just need $5^2 \cdot 7 \cdot 11 = \boxed{1925}$ to divide into $x$ . -Frestho | D | 1925 |
f86399a228ed2b83a1ca1c3861e09ae2 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_18 | Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$ . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?
$\textbf{(A)} \ \sqrt{2} \qquad \textbf{(B)} \ \frac{3}{2} \qquad \textbf{(C)} \ \frac{5}{3} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2$ | It can be seen that the diameter of the eighth sphere is equal to the radius of the seventh sphere by drawing out a diagram of the insides of the seventh sphere. The radius of the seventh sphere is $2+1=3$ , the radius of the eight sphere is $\boxed{32}$ | E | 32 |
3623c30dff20b1389fd6c1b16ff4f190 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72$ | [asy] //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); [/asy]
Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation
\[CX \cdot CB = CD \cdot CE\] \[CX(CX+XB) = (97-86)(97+86)\] \[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]
Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$ \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]
However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{61}.$ | D | 61 |
3623c30dff20b1389fd6c1b16ff4f190 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72$ | Let $BX = q$ $CX = p$ , and $AC$ meet the circle at $Y$ and $Z$ , with $Y$ on $AC$ . Then $AZ = AY = 86$ . Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , and that $p>13$ by the triangle inequality on $\triangle ACX$ . Thus, we get that $BC = p+q = \boxed{61}$ | D | 61 |
3623c30dff20b1389fd6c1b16ff4f190 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72$ | Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$ $11$ , and $61$ . Obviously, $x < x+y$ . In addition, by the Triangle Inequality, $BC < AB + AC$ , so $x+y < 183$ . Therefore, $x$ must equal $33$ , and $x+y$ must equal $\boxed{61}$ | D | 61 |
5312373d69cecd1471fe368f4f386b4b | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_23 | $ABCD$ is a square of side length $\sqrt{3} + 1$ . Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$ . The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$ , sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$ , where $a$ $b$ , and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a + b + c$
$\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23$ | We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$ . It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\sqrt{3}$ by $\sqrt{3} + 1$ . The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3}$
The area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\sqrt{2}$ (the segment bounded by $\overline{PA}$ and $\overline{PA'}$ ) and $\sqrt{6}$ (that bounded by $\overline{PC}$ and $\overline{PC'}$ ). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\overline{BC}$ . Call the point of intersection $E$ . From the previous paragraph, it is clear that $PE = \sqrt{3}$ and $BE = 1$ . This means $PB = 2$ , and $B$ swings back inside edge $\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$ , and so the angle of the segment is $60^{\circ}$ . Imagining the process in reverse, it is clear that the situation is the same with point $D$
The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\frac{1}{4} \pi (\sqrt{2})^2 - \frac{1}{2} \sqrt{2} \sqrt{2} = \frac{\pi}{2} - 1$ and $\frac{1}{4} \pi (\sqrt{6})^2 - \frac{1}{2} \sqrt{6} \sqrt{6} = \frac{3 \pi}{2} - 3$ . The other two segments both have area $\frac{1}{6} \pi (2)^2 - \frac{(2)^2 \sqrt{3}}{4} = \frac{2 \pi}{3} - \sqrt{3}$
The total area is therefore \[(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})\] \[= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}\] \[= \frac{10 \pi}{3} + 2\] \[= \frac{1}{3} (10 \pi + 6)\]
Since $a = 10$ $b = 6$ , and $c = 3$ , the answer is $a + b + c = 10 + 6 + 3 = \boxed{19}$ | C | 19 |
cb9b76b7a0f868b320ecb5f562bba123 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_25 | Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441$ | Suppose $f(z)=z^2+iz+1=c=a+bi$ . We look for $z$ with $\operatorname{Im}(z)>0$ such that $a,b$ are integers where $|a|, |b|\leq 10$
First, use the quadratic formula:
$z = \frac{1}{2} (-i \pm \sqrt{-1-4(1-c)}) = -\frac{i}{2} \pm \sqrt{ -\frac{5}{4} + c }$
Generally, consider the imaginary part of a radical of a complex number: $\sqrt{u}$ , where $u = v+wi = r e^{i\theta}$
$\operatorname{Im}(\sqrt{u}) = \operatorname{Im}(\pm \sqrt{r} e^{i\theta/2}) = \pm \sqrt{r} \sin(\theta/2) = \pm \sqrt{r}\sqrt{\frac{1-\cos\theta}{2}} = \pm \sqrt{\frac{r-v}{2}}$
Now let $u= -5/4 + c$ , then $v = -5/4 + a$ $w=b$ $r=\sqrt{v^2 + w^2}$
Note that $\operatorname{Im}(z)>0$ if and only if $\pm \sqrt{\frac{r-v}{2}}>\frac{1}{2}$ . The latter is true only when we take the positive sign, and that $r-v > 1/2$
or $v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2$ $w^2 > 1/4 + v$ , or $b^2 > a-1$
In other words, when $b^2 > a-1$ , the equation $f(z)=a+bi$ has unique solution $z$ in the region $\operatorname{Im}(z)>0$ ; and when $b^2 \leq a-1$ there is no solution. Therefore the number of desired solution $z$ is the same as the number of ordered pairs $(a,b)$ such that integers $|a|, |b|\leq 10$ , and that $b^2 \geq a$
When $a\leq 0$ , there is no restriction on $b$ so there are $11\cdot 21 = 231$ pairs;
when $a > 0$ , there are $2(1+4+9+10+10+10+10+10+10+10)=2(84)=168$ pairs.
So there are $231+168=\boxed{399}$ in total. | null | 399 |
cb9b76b7a0f868b320ecb5f562bba123 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_25 | Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441$ | We consider the function $f(z)$ as a mapping from the 2-D complex plane onto itself. We complete the square of $f(z)=z^2+iz+1=(z+\frac{i}{2})^2+\frac{5}{4}$
Now, we must decide the range of $f(z)$ based on the domain of $z$ $\operatorname{Im}(z)>0$ . To do this, we are interested in mapping the boundary line $\operatorname{Im}(z)=0$ . To make the mapping simpler, let $f(z)=g(z)+\frac{5}{4}$ , or $g(z)=(z+\frac{i}{2})^2$
We intend to map of the line $\operatorname{Im}(z)=0$ using the function $g(z)$ . This transformation is equivalent to the polar equation $r=(\frac{1}{2}\csc(\frac{\theta}{2}))^2$ . Using polar and trig identities, we can restate this equation as the rectangular form of a parabola,
$x=y^2-\frac{1}{4}$
where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$ . So, we conclude that $f(z)$ maps the line $\operatorname{Im}(z)=0$ to the parabola
$x=y^2-\frac{1}{4}+\frac{5}{4}=y^2+1$
A quick check reveals that the range of $f(z)$ is to the left of the parabola, meaning that any point on or to the right of parabola cannot be reached.
Since the problem requires $|\operatorname{Re}(z)|$ and $|\operatorname{Im}(z)|$ to both be integers and at most 10, all that remains is counting all points with integer coordinates in the range of $f(z), \operatorname{Im}(z)>0$ . To do this, we employ complementary counting.
The points of interest are $|\operatorname{Re}(z)|\leq 10$ and $|\operatorname{Im}(z)|\leq 10$ , resulting in a total of $441$ points. For lattice points on or to the right of the parabola, there are $10$ points for $x=0$ $9$ points for $x=\pm 1$ $6$ points for $x=\pm 2$ , and $1$ point for $x=\pm 3$ . Summing it all together, our answer is $441-(10+2*9+2*6+2*1)=\boxed{399}$ | null | 399 |
d25440e037547713efc86a480d1243d0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_1 | On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11$ | Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$ | C | 5 |
d6df558445f7b99111f769078b661a9f | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_2 | Mr. Green measures his rectangular garden by walking two of the sides and finds that it is $15$ steps by $20$ steps. Each of Mr. Green's steps is $2$ feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?
$\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400$ | Since each step is $2$ feet, his garden is $30$ by $40$ feet. Thus, the area of $30(40) = 1200$ square feet. Since he is expecting $\frac{1}{2}$ of a pound per square foot, the total amount of potatoes expected is $1200 \times \frac{1}{2} = \boxed{600}$ | A | 600 |
b4bbc34b18237def72a199d9d1591ff9 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_3 | When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$ | Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\boxed{149}$ | D | 149 |
caf827e233080ed25f992b7c1311b098 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$ | Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$ | B | 16 |
caf827e233080ed25f992b7c1311b098 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$ | Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\] | B | 16 |
caf827e233080ed25f992b7c1311b098 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$ | Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac{m}{40} + \frac{m}{10} = \frac{m}{8}$ . The total distance that both drive is equal to $2m$ so the total miles per gallon can be expressed as $\frac{2m}{\frac{m}{8}} = \boxed{16}$ | B | 16 |
7f035960235766a9df7c38da312eb7f2 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_5 | The average age of $33$ fifth-graders is $11$ . The average age of $55$ of their parents is $33$ . What is the average age of all of these parents and fifth-graders?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23.25 \qquad \textbf{(C)}\ 24.75 \qquad \textbf{(D)}\ 26.25 \qquad \textbf{(E)}\ 28$ | The sum of the ages of the fifth graders is $33 * 11$ , while the sum of the ages of the parents is $55 * 33$ . Therefore, the total sum of all their ages must be $2178$ , and given $33 + 55 = 88$ people in total, their average age is $\frac{2178}{88} = \frac{99}{4} = \boxed{24.75}$ | C | 24.75 |
2e65a1383ada9da1400fafcce7f5eb74 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_6 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$ . Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$ . We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$ . Notice this is a circle with radius $0$ , which only contains one point. So, the only point is $(5, -3)$ , so the sum is $5 + (-3) = 2 \implies \boxed{2}$ . ~ asdf334 | B | 2 |
2e65a1383ada9da1400fafcce7f5eb74 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_6 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ . Equality holds when the value inside the parhentheses is equal to $0$ . We find that \[(x,y) = (5,-3)\] and the sum we are looking for is \[5+(-3)=2 \implies \boxed{2}.\] - Honestly | B | 2 |
14d9eebd3c1079e141f6b562fd313978 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$ | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$ . Since we are starting from 1 every turn, the 53rd number said will be $\boxed{8}$ | E | 8 |
14d9eebd3c1079e141f6b562fd313978 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$ | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n(n+1)}{2}$ under 53, we can easily see that if we double this it's $n^2 + n \simeq 106$ , and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, $\frac{9(9+1)}{2} = 45$ and $53 - 45 = \boxed{8}$ | E | 8 |
14d9eebd3c1079e141f6b562fd313978 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$ | Let $T(n)$ denote the $n$ th triangle number. Then, observe that the $T(n)$ th number said is $n$ . It follows that the $55$ th number is $10$ (as $55 = T(10)$ ). Thus, the $53$ rd number is $10 - 2 = 8$ , which is answer choice $\boxed{8}$ | E | 8 |
98e42292a58b1ba8034ab5610b8877e1 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_9 | What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | Looking at the prime numbers under $12$ , we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of $2$ $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$ , and $\left\lfloor\frac{12}{5}\right\rfloor=2$ factors of $5$ . All greater primes are represented once or none in $12!$ , so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$ . Therefore, the prime factorization of the square is $2^{10}\cdot3^4\cdot5^2$ . To find the square root of this, we halve the exponents, leaving $2^5\cdot3^2\cdot5$ . The sum of the exponents is $\boxed{8}$ | C | 8 |
fdbafaf9e409e9045c49e0625546e111 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_10 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$ | If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.
In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.
Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\boxed{103}$ silver tokens, at which point it is clear he cannot use the booths anymore. | E | 103 |
fdbafaf9e409e9045c49e0625546e111 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_10 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$ | We can approach this problem by assuming he goes to the red booth first. You start with $75 \text{R}$ and $75 \text{B}$ and at the end of the first booth, you will have $1 \text{R}$ and $112 \text{B}$ and $37 \text{S}$ . We now move to the blue booth, and working through each booth until we have none left, we will end up with: $1 \text{R}$ $2 \text{B}$ and $103 \text{S}$ . So, the answer is $\boxed{103}$ | E | 103 |
fdbafaf9e409e9045c49e0625546e111 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_10 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?
$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$ | Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \[R(x,y)=-2x+y+75\] \[B(x,y)=x-3y+75\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins, namely when he has a red coin and $2$ blue coins. We can then create a system of equations: \[1=-2x+y+75\] \[2=x-3y+75\] Solving yields $x=59$ and $y=44$ . Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\boxed{103}$ silver coins in total. | E | 103 |
d812b4129488b2277e42f679cf7267bd | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_12 | Cities $A$ $B$ $C$ $D$ , and $E$ are connected by roads $\widetilde{AB}$ $\widetilde{AD}$ $\widetilde{AE}$ $\widetilde{BC}$ $\widetilde{BD}$ $\widetilde{CD}$ , and $\widetilde{DE}$ . How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.)
[asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (C); dot (D); dot (E); label("$A$",A,S); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,W); guide squiggly(path g, real stepsize, real slope=45) { real len = arclength(g); real step = len / round(len / stepsize); guide squig; for (real u = 0; u < len; u += step){ real a = arctime(g, u); real b = arctime(g, u + step / 2); pair p = point(g, a); pair q = point(g, b); pair np = unit( rotate(slope) * dir(g,a)); pair nq = unit( rotate(0 - slope) * dir(g,b)); squig = squig .. p{np} .. q{nq}; } squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))}; return squig; } pen pp = defaultpen + 2.718; draw(squiggly(A--B, 4.04, 30), pp); draw(squiggly(A--D, 7.777, 20), pp); draw(squiggly(A--E, 5.050, 15), pp); draw(squiggly(B--C, 5.050, 15), pp); draw(squiggly(B--D, 4.04, 20), pp); draw(squiggly(C--D, 2.718, 20), pp); draw(squiggly(D--E, 2.718, -60), pp); [/asy]
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$ | Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$ , there is only one possible path to take out of cities $C$ or $E$ .
So the diagram is as follows:
[asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (D); label("$A$",A,S); label("$B$",B,SE); label("$D$",D,N); draw(A--B..D..cycle); draw(A--D); draw(B--D); [/asy]
Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$ $D$ to $A$ $B$ to $D$ and $D$ to $B$ .:
Case 1 $A \Rightarrow D$ : From $D$ , if the path returns to $A$ , then the next path must go to $B\Rightarrow D \Rightarrow B$ . There are $2 \cdot 1 \cdot 2 = 4$ possibilities of the path $ADABDB$ . If the path goes to $D$ from $B$ , then the path must continue with either $BDAB$ or $BADB$ . There are $2 \cdot 2 \cdot 2 = 8$ possibilities. So, this case gives $4+8=12$ different possibilities.
Case 2 $A \Rightarrow B$ : The path must continue with $BDADB$ . There are $2 \cdot 2 = 4$ possibilities for this case.
Putting the two cases together gives $12+4 = \boxed{16}$ | D | 16 |
e97f4e80dcac51a806503d76abc5c4a0 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_13 | The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$ . Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD$
$\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250$ | Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ $a+d$ $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$
For the sake of simplicity, lets rename the angles of each similar triangle. Let $\angle ADB = \angle CBD = \alpha$ $\angle DBA = \angle DCB = \beta$ $\angle CDB = \angle BAD = \gamma$
Now the four angles of $ABCD$ are $\beta$ $\alpha + \beta$ $\gamma$ , and $\alpha + \gamma$
As for the similar triangles, we can name their angles $y$ $y+b$ , and $y+2b$ . Therefore $y+y+b+y+2b=180$ and $y+b=60$ . Because these 3 angles are each equal to one of $\alpha, \beta, \gamma$ , we know that one of these three angles is equal to 60 degrees.
Now we we use trial and error. Let $\alpha = 60^{\circ}$ . Then the angles of ABCD are $\beta$ $60^{\circ} + \beta$ $\gamma$ , and $60^{\circ} + \gamma$ . Since these four angles add up to 360, then $\beta + \gamma= 120$ . If we list them in increasing value, we get $\beta$ $\gamma$ $60^{\circ} + \beta$ $60^{\circ} + \gamma$ . Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, $\alpha, \beta, \gamma$ also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.
If we apply the same reasoning to $\beta$ and $\gamma$ , we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, $\boxed{240}$ is the correct answer. | D | 240 |
1a9c78ee9f445af92454bc5ff7290bb3 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_14 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$ | Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, let one of them be $0$ WLOG , assume that $x_{1} = 0$ . Thus, the smallest possible value of $y_{1}$ is $8$ ; and since the sequences are non-decreasing we get $y_{2} \ge 8$ . To minimize, let $y_{2} = 8$ . Thus, $5y_{1} + 8y_{2} = 40 + 64 = \boxed{104}$ | C | 104 |
1a9c78ee9f445af92454bc5ff7290bb3 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_14 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$ | WLOG, let $a_i$ $b_i$ be the sequences with $a_1<b_1$ . Then \[N=5a_1+8a_2=5b_1+8b_2\] or \[5a_1+8a_2=5(a_1+c)+8(a_2-d)\] for some natural numbers $c$ $d$ . Thus $5c=8d$ . To minimize $c$ and $d$ , we have $(c,d)=(8,5)$ , or \[5a_1+8a_2=5(a_1+8)+8(a_2-5).\] To minimize $a_1$ and $b_1$ , we have $(a_1,b_1)=(0,0+c)=(0,8)$ . Using the same method, since $b_2\ge b_1$ , we have $b_2\ge8$
Thus the minimum $N=5b_1+8b_2=104\boxed{104}$ | C | 104 |
f947d416ed5e04489c729d7003bd1fe9 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_15 | The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead to an extra factor in the numerator. The highest prime less than $61$ is $59$ , so there must be a factor of $59$ in the denominator. It follows that $b_1 = 59$ (to minimize $b_1$ as well), so the answer is $|61-59| = \boxed{2}.\] | B | 2 |
91e8013016ba83d97f7f2f792683efb5 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_16 | Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$ . The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$ | The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$ , and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure $180^\circ - 108^\circ = 72^\circ$ . The base angles are equal, so the triangles must be isosceles.
Let one of the sides of the pentagon have length $x_1$ (and the others $x_2, x_3, x_4, x_5$ ). Then, by trigonometry, the non-base sides of the triangle sitting on that side of the pentagon each has length $\frac{x_1}{2} \sec 72^\circ$ , and so the two sides together have length $x_1 \sec 72^\circ$ . To find the perimeter of the star, we sum up the lengths of the non-base sides for each of the five triangles to get $(x_1+x_2+x_3+x_4+x_5) \sec 72^\circ = (1) \sec 72^\circ = \sec 72^\circ$ (because the perimeter of the pentagon is $1$ ). The perimeter of the star is constant, so the difference between the maximum and minimum perimeters is $\boxed{0}$ | A | 0 |
91e8013016ba83d97f7f2f792683efb5 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_16 | Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$ . The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$ | The extreme case, that results in the minimum and/or maximum, would probably be a pentagon that approaches a degenerate pentagon. However, due to the way the problem is phrased, we know there exists a minimum and maximum; therefore, we can reasonably assume that the star's perimeter is constant, and answer with $\boxed{0}$ | A | 0 |
359db4c425d3de5bd0d82e85d75bd845 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is $\tfrac{12}{15} = \tfrac{4}{5}$ , and the ratio of the shorter leg to the hypotenuse is $\tfrac{9}{15} = \tfrac{3}{5}$ . It follows that $AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)$
Let $x=DF$ . By Ptolemy's Theorem , we have \[13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.\] Dividing by $13$ we get $x+4=7.2\implies x=\frac{16}{5}$ so our answer is $\boxed{21}$ | B | 21 |
359db4c425d3de5bd0d82e85d75bd845 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | From solution 1, we know that $AD = 12$ and $DC = 9$ . Since $\triangle ADC \sim \triangle DEC$ , we can figure out that $DE = \frac{36}{5}$ . We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°. Therefore, $\angle{B} = 180 - \angle{DFA} = \angle{EFA}$ , and triangles $\triangle AEF \sim \triangle ADB$ . Solving the resulting proportion gives $EF = 4$ . Therefore, $DF = ED - EF = \frac{16}{5}$ $m + n = 16 + 5 = 21$ and our answer is $\boxed{21}$ | B | 21 |
359db4c425d3de5bd0d82e85d75bd845 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Using the Pythagorean theorem, we now get \[BF = \sqrt{\left(\frac45x+ 5\right)^2 + \left(\frac35x\right)^2}\] and $AF$ can be found out noting that $AE$ is just $\tfrac{48}5$ through base times height (since $12\cdot 9 = 15 \cdot \tfrac{36}5$ , similar triangles gives $AE = \tfrac{48}5$ ), and that $EF$ is just $\tfrac{36}5 - x$ . From there, \[AF = \sqrt{\left(\frac{36}5 - x\right)^2 + \left(\frac{48}5\right)^2}.\] Now, $BF^2 + AF^2 = 169$ , and squaring and adding both sides and subtracting a 169 from both sides gives $2x^2 - \tfrac{32}5x = 0$ , so $x = \tfrac{16}5$ . Thus, the answer is $\boxed{21}$ | B | 21 |
359db4c425d3de5bd0d82e85d75bd845 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30$ | First, we find $BD = 5$ $DC = 9$ , and $AD = 12$ via the Pythagorean Theorem or by using similar triangles. Next, because $DE$ is an altitude of triangle $ADC$ $DE = \frac{AD\cdot DC}{AC} = \frac{36}{5}$ . Using that, we can use the Pythagorean Theorem and similar triangles to find $EC = \frac{27}{5}$ and $AE = \frac{48}{5}$
Points $A$ $B$ $D$ , and $F$ all lie on a circle whose diameter is $AB$ . Let the point where the circle intersects $AC$ be $G$ . Using power of a point, we can write the following equation to solve for $AG$ \[DC\cdot BC = CG\cdot AC\] \[9\cdot 14 = CG\cdot 15\] \[CG = 126/15\] Using that, we can find $AG = \frac{99}{15}$ , and using $AG$ , we can find that $GE = 3$
We can use power of a point again to solve for $DF$ \[FE\cdot DE = GE\cdot AE\] \[(\frac{36}{5} - DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}\] \[\frac{36}{5} - DF = 4\] \[DF = \frac{16}{5} = \frac{m}{n}\] Thus, $m+n = 16+5 = 21$ $\boxed{21}$ | B | 21 |
232a05179c51dffc1021c5a1a585187a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_21 | Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form $y=ax+b$ with $a$ and $b$ integers such that $a\in \{-2,-1,0,1,2\}$ and $b\in \{-3,-2,-1,1,2,3\}$ . No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D)}\ 840\qquad\textbf{(E)}\ 870$ | Being on two parabolas means having the same distance from the common focus and both directrices. In particular, you have to be on an angle bisector of the directrices, and clearly on the same "side" of the directrices as the focus. So it's easy to see there are at most two solutions per pair of parabolae. Convexity and continuity imply exactly two solutions unless the directrices are parallel and on the same side of the focus.
So out of $2\dbinom{30}{2}$ possible intersection points, only $2*5*2*\dbinom{3}{2}$ fail to exist. This leaves $870-60=810=\boxed{810}$ solutions. | C | 810 |
232a05179c51dffc1021c5a1a585187a | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_21 | Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form $y=ax+b$ with $a$ and $b$ integers such that $a\in \{-2,-1,0,1,2\}$ and $b\in \{-3,-2,-1,1,2,3\}$ . No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D)}\ 840\qquad\textbf{(E)}\ 870$ | Through similar reasoning as above in Solution I, determine that two parabolas that have a common focus intersect zero times if there directrixes are parallel and the focus lies on the same side of both directrixes, and intersect twice otherwise. Thereby, as each parabola will intersect $30-3 = 27$ other parabolas twice, we see that the answer is \[2 \times \frac{30 \times 27}{2} = \boxed{810}.\] | null | 810 |
c4ebcf96c8ffc2e91ec1b414b9523541 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_22 | Let $m>1$ and $n>1$ be integers. Suppose that the product of the solutions for $x$ of the equation
\[8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0\]
is the smallest possible integer. What is $m+n$
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272$ | Rearranging logs, the original equation becomes
\[\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0\]
By Vieta's Theorem, the sum of the possible values of $\log x$ is $\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt[8]{m^7n^6}$ . But the sum of the possible values of $\log x$ is the logarithm of the product of the possible values of $x$ . Thus the product of the possible values of $x$ is equal to $\sqrt[8]{m^7n^6}$
It remains to minimize the integer value of $\sqrt[8]{m^7n^6}$ . Since $m, n>1$ , we can check that $m = 2^2$ and $n = 2^3$ work. Thus the answer is $4+8 = \boxed{12}$ | A | 12 |
b821095c35cc0cb2f237730a266a7fcc | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_23 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that $N \equiv a \pmod{6}$
also that $N \equiv b \pmod{5}$
Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b < 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$ $N \equiv a \pmod{6}$ $N \equiv a \pmod{5}$ $\implies N=a \pmod{30}$ $0 \le a \le 4$
Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$
Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$ , ;
Also, we have already found which digits of $y$ will add up into the units digits of $2N$
Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work)
Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation. \[N \equiv 30x \pmod{36}\] \[N \equiv 30x \equiv 5x \pmod{25}\] Both of those must add up to \[2N\equiv60x \pmod{100}\]
$33 \ge x \ge 4$
Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. \[N \equiv 5x \pmod{6}\] \[N \equiv 6x \equiv x \pmod{5}\] \[2N\equiv 6x \pmod{10}\]
Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x)
Then
\[N \equiv 5m+n \pmod{5}\] \[N \equiv 25m+5n \pmod{6}\] \[2N\equiv30m + 6n \pmod{10}\]
and this simplifies to
\[N \equiv n \pmod{5}\] \[N \equiv m+5n \pmod{6}\] \[2N\equiv 6n \pmod{10}\]
From careful inspection, this is true when
$n=0, m=6$
$n=1, m=6$
$n=2, m=2$
$n=3, m=2$
$n=4, m=4$
This gives you $5$ choices for $x$ , and $5$ choices for $y$ , so the answer is $5* 5 = \boxed{25}$ | E | 25 |
b821095c35cc0cb2f237730a266a7fcc | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_23 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | Notice that there are exactly $1000-100=900=5^2\cdot 6^2$ possible values of $N$ . This means, in $100\le N\le 999$ , every possible combination of $2$ digits will happen exactly once. We know that $N=900,901,902,903,904$ works because $900\equiv\dots00_5\equiv\dots00_6$
We know for sure that the units digit will add perfectly every $30$ added or subtracted, because $\text{lcm }5,6=30$ . So we only have to care about cases of $N$ every $30$ subtracted. In each case, $2N$ subtracts $6$ /adds $4$ $N_5$ subtracts $1$ and $N_6$ adds $1$ for the $10$ 's digit.
\[\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}\]
\[\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}\]
\[\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}\]
As we can see, there are $5$ cases, including the original, that work. These are highlighted in $\textcolor{red}{\text{red}}$ . So, thus, there are $5$ possibilities for each case, and $5\cdot 5=\boxed{25}$ | E | 25 |
b821095c35cc0cb2f237730a266a7fcc | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_23 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | Notice that $N_5$ ranges from $3$ to $5$ digits and $N_6$ ranges from $3$ to $4$ digits.
Then let $a_i$ $b_i$ denotes the digits of $N_5$ $N_6$ , respectively such that \[0\le a_i<5,0\le b_i<6\] Thus we have \[N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4\] \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Now we are given \[2N \equiv S \equiv N_5+N_6\pmod{100}\] \[2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}\] \[1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}\] \[50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}\] Canceling out $a_5$ left with \[50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}\]
Since $a_5$ $b_4$ determine the unit digits of the two sides of the congruence equation, we have $a_5=b_4=0,1,2,3,4$ . Thus,
\[50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}\] canceling out $10a_4$ , we have \[50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}\] \[5a_1+5a_2+5a_3 \equiv b_3\pmod{10}\] \[5(a_1+a_2+a_3) \equiv b_3\pmod{10}\]
Thus $b_3$ is a multiple of $5$
Now going back to our original equation \[625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4\] Since $a_5=b_4$ \[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\] \[5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)\] \[5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]\]
Since the left side is a multiple of $5$ , then so does the right side. Thus $5\mid6(6b_1+b_2)+b_3$
Since we already know that $5\mid b_3$ , then $5\mid6(6b_1+b_2)$ , from where we also know that $5\mid6b_1+b_2$
For $b_1,b_2<6$ , there is a total of 7 ordered pairs that satisfy the condition. Namely,
\[(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)\]
Since $N_6$ has at least $3$ digits, $(b_1,b_2)=(0,0)$ doesn't work. Furthermore, when $b_1=5$ $216b_1$ exceeds $1000$ which is not possible as $N$ is a three digit number, thus $(b_1,b_2)=(5,0)$ won't work as well.
Since we know that $a_i<5$ , for each of the ordered pairs $(b_1,b_2)$ , there is respectively one and only one solution $(a_1,a_2,a_3,a_4)$ that satisfies the equation
\[625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3\]
Thus there are five solutions to the equation. Also since we have 5 possibilities for $a_5=b_4$ , we have a total of $5\cdot5=25$ values for $N$ $\boxed{25}$ | E | 25 |
b821095c35cc0cb2f237730a266a7fcc | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_23 | Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$ . For example, if $N = 749$ , Bernardo writes the numbers $10,\!444$ and $3,\!245$ , and LeRoy obtains the sum $S = 13,\!689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ | Observe that the maximum possible value of the sum of the last two digits of the base $5$ number and the base $6$ number is $44+55=99$ .
Let $N \equiv a \pmod {25}$ and $N \equiv b \pmod {36}$
If $a < \frac{25}{2}$ $2N \equiv 2a \pmod {25}$ and if $a > \frac{25}{2}$ $2N \equiv 2a - 25 \pmod {25}$
Using the same logic for $b$ , if $b < 18$ $2N \equiv 2b \pmod {36}$ , and in the other case $2N \equiv 2b - 36 \pmod {36}$
We can do four cases:
Case 1: $a + b = 2a - 25 + 2b - 36 \implies a + b = 61$
For this case, there is trivially only one possible solution, $(a, b) = (25, 36)$ , which is equivalent to $(a, b) = (0, 0)$
Case 2: $a + b = 2a - 25 + 2b \implies a + b = 25$
Note that in this case, $a \geq 13$ must hold, and $b < 18$ must hold.
We find the possible ordered pairs to be: $(13, 12), (14, 11), (15, 10), ..., (24, 1)$ for a total of $12$ ordered pairs.
Case 3: $a + b = 2a + 2b - 36 \implies a + b = 36$
Note that in this case, $b \geq 18$ must hold, and $a < 13$ must hold.
We find the possible ordered pairs to be: $(24, 12), (25, 11), (26, 10), ..., (35, 1)$ for a total of $12$ ordered pairs.
Case 4: $a + b = 2a + 2b$
Trivially no solutions except $(a, b) = (0, 0)$ , which matches the solution in Case 1, which makes this an overcount.
By CRT, each solution $(a, b)$ corresponds exactly one positive integer in a set of exactly $\text{lcm} (25, 36) = 900$ consecutive positive integers, and since there are $900$ positive integers between $100$ and $999$ , our induction is complete, and our answer is $1 + 12 + 12 = \boxed{25}$ | E | 25 |
8ee93da252f26e19e65eb6b45f19c70e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_25 | Let $G$ be the set of polynomials of the form \[P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,\] where $c_1,c_2,\cdots, c_{n-1}$ are integers and $P(z)$ has distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$
$\textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056$ | If we factor into irreducible polynomials (in $\mathbb{Q}[x]$ ), each factor $f_i$ has exponent $1$ in the factorization and degree at most $2$ (since the $a+bi$ with $b\ne0$ come in conjugate pairs with product $a^2+b^2$ ). Clearly we want the product of constant terms of these polynomials to equal $50$ ; for $d\mid 50$ , let $f(d)$ be the number of permitted $f_i$ with constant term $d$ . It's easy to compute $f(1)=2$ $f(2)=3$ $f(5)=5$ $f(10)=5$ $f(25)=6$ $f(50)=7$ , and obviously $f(d) = 1$ for negative $d\mid 50$
Note that by the distinctness condition, the only constant terms $d$ that can be repeated are those with $d^2\mid 50$ and $f(d)>1$ , i.e. $+1$ and $+5$ . Also, the $+1$ s don't affect the product, so we can simply count the number of polynomials with no constant terms of $+1$ and multiply by $2^{f(1)} = 4$ at the end.
We do casework on the (unique) even constant term $d\in\{\pm2,\pm10,\pm50\}$ in our product. For convenience, let $F(d)$ be the number of ways to get a product of $50/d$ without using $\pm 1$ (so only using $\pm5,\pm25$ ) and recall $f(-1) = 1$ ; then our final answer will be $2^{f(1)}\sum_{d\in\{2,10,50\}}(f(-d)+f(d))(F(-d)+F(d))$ . It's easy to compute $F(-50)=0$ $F(50)=1$ $F(-10)=f(5)=5$ $F(10)=f(-5)=1$ $F(-2)=f(-25)+f(-5)f(5)=6$ $F(2)=f(25)+\binom{f(5)}{2}=16$ , so we get \[4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = \boxed{528}\] | B | 528 |
8ee93da252f26e19e65eb6b45f19c70e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_25 | Let $G$ be the set of polynomials of the form \[P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,\] where $c_1,c_2,\cdots, c_{n-1}$ are integers and $P(z)$ has distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$
$\textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056$ | Disregard sign; we can tack on $x-1$ if the product ends up being negative.
$1: \pm i,-1$ (2) (1 is not included)
$2: \pm 2, \pm 1\pm i$ (4)
$5: \pm 2\pm i, \pm 1\pm 2i, \pm 5$ (6)
$10: \pm 3\pm i, \pm 1\pm 3i, \pm 10$ (6)
$25: \pm 25, \pm 3\pm 4i, \pm 4\pm 3i, \pm 5i$ (7)
$50: \pm 50, \pm 1\pm 7i, \pm7\pm i, \pm 5\pm 5i$ (8)
Our answer is $2^2\left(4\cdot\binom{6}{2}+6\cdot 6+4\cdot 7+8\right)=\boxed{528.}$ | null | 528. |
8ee93da252f26e19e65eb6b45f19c70e | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_25 | Let $G$ be the set of polynomials of the form \[P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,\] where $c_1,c_2,\cdots, c_{n-1}$ are integers and $P(z)$ has distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$
$\textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056$ | By Vieta's formula $50$ is the product of all $n$ roots. As the roots are all in the form $a + bi$ , there must exist a conjugate $a-bi$ for each root.
$(a+bi)(a-bi) = a^2 + b^2$
$50 = 2 \cdot 5^2$
If $a \neq b \neq 0$ , the roots can be $a \pm bi$ $-a \pm bi$ $b \pm ai$ $-b \pm ai$ , totaling $4$ pairs of roots.
If $a = b$ , the roots can be $a \pm ai$ $-a \pm ai$ , totaling $2$ pairs of roots.
If $a \neq b$ $b = 0$ , the roots can be $\pm a$ $\pm ai$ , totaling $2$ pairs of roots.
\begin{align*} 2 \cdot 25 &= (1^2+1^2)5^2 &: 2 \cdot 2 = 4\\ 2 \cdot 25 &= 2 \cdot 5^2 &: 2 \cdot 2 = 4\\ 2 \cdot 25 &= (1^2+1^2) \cdot (3^2+4^2) &: 2 \cdot 4 = 8\\ 2 \cdot 25 &= 2 \cdot (3^2+4^2) &: 2 \cdot 4 = 8 \end{align*}
\begin{align*} 10 \cdot 5 &= (1^2+3^2)(1^2+2^2) &&: 4 \cdot 4 = 16\\ 10 \cdot 5 &= 10 \cdot (1^2+2^2) &&: 2 \cdot 4 = 8\\ 10 \cdot 5 &= (1^2+3^2) \cdot 5 &&: 4 \cdot 2 = 8\\ 10 \cdot 5 &= 10 \cdot 5 &&: 2 \cdot 2 = 4\\ \end{align*}
\begin{align*} 2 \cdot 5 \cdot 5&= (1^2+1^2)(1^2+2^2)(1^2+2^2) &&: 2 \cdot 4 \cdot 4 = 32\\ 2 \cdot 5 \cdot 5&= 2 \cdot (1^2+2^2)(1^2+2^2) &&: 2 \cdot 4 \cdot 4 = 32\\ 2 \cdot 5 \cdot 5&= 2 \cdot 5 \cdot (1^2+2^2) &&: 2 \cdot 2 \cdot 4 = 16\\ 2 \cdot 5 \cdot 5&= 2 \cdot 5 \cdot 5 &&: 2 \cdot 2 \cdot 2 = 8\\ 2 \cdot 5 \cdot 5&= (1^2+1^2) \cdot 5 \cdot (1^2+2^2) &&: 2 \cdot 2 \cdot 4 = 16\\ 2 \cdot 5 \cdot 5&= (1^2+1^2) \cdot 5 \cdot 5 &&: 2 \cdot 2 \cdot 2 = 8\\ \end{align*}
\begin{align*} (1^2+7^2) &: 4\\ (5^2+5^2) &: 2\\ 50 &: 2 \end{align*}
$4+4+8+8+16+8+8+4+32+32+16+8+16+8+4+2+2 = 176$
For each case $1^2$ can be added, yielding 2 more cases $(\pm 1, \pm i)$ $176 \cdot 3 = \boxed{528}$ | B | 528 |
2b90cde9731072c96e3e17b9a3ada5d2 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_1 | A bug crawls along a number line, starting at $-2$ . It crawls to $-6$ , then turns around and crawls to $5$ . How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | [asy] draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); [/asy]
Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{15}$ | E | 15 |
4e7242b4d6af59ae21c1718730c1a624 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_2 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | Cagney can frost one in $20$ seconds, and Lacey can frost one in $30$ seconds. Working together, they can frost one in $\frac{20\cdot30}{20+30} = \frac{600}{50} = 12$ seconds. In $300$ seconds ( $5$ minutes), they can frost $\boxed{25}$ cupcakes. | D | 25 |
4e7242b4d6af59ae21c1718730c1a624 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_2 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | In $300$ seconds ( $5$ minutes), Cagney will frost $\dfrac{300}{20} = 15$ cupcakes, and Lacey will frost $\dfrac{300}{30} = 10$ cupcakes. Therefore, working together they will frost $15 + 10 = \boxed{25}$ cupcakes. | D | 25 |
4e7242b4d6af59ae21c1718730c1a624 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_2 | Cagney can frost a cupcake every 20 seconds and Lacey can frost a cupcake every 30 seconds. Working together, how many cupcakes can they frost in 5 minutes?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 30$ | Since Cagney frosts $3$ cupcakes a minute, and Lacey frosts $2$ cupcakes a minute, they together frost $3+2=5$ cupcakes a minute. Therefore, in $5$ minutes, they frost $5\times5 = 25 \Rightarrow \boxed{25}$ | D | 25 |
cf884977c4c93579e9abc6bd71e4643d | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_3 | A box $2$ centimeters high, $3$ centimeters wide, and $5$ centimeters long can hold $40$ grams of clay. A second box with twice the height, three times the width, and the same length as the first box can hold $n$ grams of clay. What is $n$
$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 160\qquad\textbf{(C)}\ 200\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 280$ | The first box has volume $2\times3\times5=30\text{ cm}^3$ , and the second has volume $(2\times2)\times(3\times3)\times(5)=180\text{ cm}^3$ . The second has a volume that is $6$ times greater, so it holds $6\times40=\boxed{240}$ grams. | D | 240 |
5db163ce47d9cd140524ef8966d57594 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_5 | A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$ | So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$
Observe that since there are $280$ pieces of fruit, \[b+r+g+c=280.\]
Since there are twice as many raspberries as blueberries, \[2b=r.\]
The fact that there are three times as many grapes as cherries implies, \[3c=g.\]
Because there are four times as many cherries as raspberries, we deduce the following: \[4r=c.\]
Note that we are looking for $c.$ So, we try to rewrite all of the other variables in terms of $c.$ The third equation gives us the value of $g$ in terms of $c$ already. We divide the fourth equation by $4$ to get that $r=\frac{c}{4}.$ Finally, substituting this value of $r$ into the first equation provides us with the equation $b=\frac{c}{8}$ and substituting yields: \[\frac{c}{4}+\frac{c}{8}+3c+c=280\] Multiply this equation by $8$ to get: \[2c+c+24c+8c=8\cdot 280,\] \[35c=8\cdot 280,\] \[c=64.\] \[\boxed{64}\] | D | 64 |
4f6509984ec118e7e5d671b12407efea | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_6 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:
$a+b=12$
$b+c=17$
$a+c=19$
Adding these equations together, we get that
$2(a+b+c)=48$ and
$a+b+c=24$
Substituting the original equations into this one, we find
$c+12=24$
$a+17=24$
$b+19=24$
Therefore, our numbers are 12, 7, and 5. The middle number is $\boxed{7}$ | D | 7 |
4f6509984ec118e7e5d671b12407efea | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_6 | The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:
$a+b=12$
$a+c=17$
$b+c=19$
To isolate $b$ , We add the first and last equations and then subtract the second one.
$(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7$
Because $b$ is the middle number, the middle number is $\boxed{7}$ | D | 7 |
ae7f375c48b22c157dae0b8b19fb82bf | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_7 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $12*\frac{a_1+a_{12}}{2}=360$ , which leads us to $a_1 + a_{12} = 60$ $a_{12}$ , the largest term of the progression, can also be expressed as $a_1+11d$ , where $d$ is the common difference. Since each angle measure must be an integer, $d$ must also be an integer. We can isolate $d$ by subtracting $a_1$ from $a_{12}$ like so: $a_{12}-a_1=a_1+11d-a_1=11d$ . Since $d$ is an integer, the difference between the first and last terms, $11d$ , must be divisible by $11.$ Since the total difference must be less than $60$ , we can start checking multiples of $11$ less than $60$ for the total difference between $a_1$ and $a_{12}$ . We start with the largest multiple, because the maximum difference will result in the minimum value of the first term. If the difference is $55$ $a_1=\frac{60-55}{2}=2.5$ , which is not an integer, nor is it one of the five options given. If the difference is $44$ $a_1=\frac{60-44}{2}$ , or $\boxed{8}$ | C | 8 |
ae7f375c48b22c157dae0b8b19fb82bf | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_7 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | If we let $a$ be the smallest sector angle and $r$ be the difference between consecutive sector angles, then we have the angles $a, a+r, a+2r, \cdots. a+11r$ . Use the formula for the sum of an arithmetic sequence and set it equal to 360, the number of degrees in a circle.
\begin{align*} \frac{a+a+11r}{2}\cdot 12 &= 360\\ 2a+11r &= 60\\ a &= \frac{60-11r}{2} \end{align*}
All sector angles are integers so $r$ must be a multiple of 2. Plug in even integers for $r$ starting from 2 to minimize $a.$ We find this value to be 4 and the minimum value of $a$ to be $\frac{60-11(4)}{2} = \boxed{8}$ | C | 8 |
ae7f375c48b22c157dae0b8b19fb82bf | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_7 | Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | Starting with the smallest term, $a - 5x \cdots a, a + x \cdots a + 6x$ where $a$ is the sixth term and $x$ is the difference. The sum becomes $12a + 6x = 360$ since there are $360$ degrees in the central angle of the circle. The only condition left is that the smallest term in greater than zero. Therefore, $a - 5x > 0$ \[2a + x = 60\] \[x = 60 - 2a\] \[a - 5(60 - 2a) > 0\] \[11a > 300\] Since $a$ is an integer, it must be $28$ , and therefore, $x$ is $4$ $a - 5x$ is $\boxed{8}$ | C | 8 |
86fbefa4326a4714bca6011753c40117 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_10 | A triangle has area $30$ , one side of length $10$ , and the median to that side of length $9$ . Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$
$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$ | $AB$ is the side of length $10$ , and $CD$ is the median of length $9$ . The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$ . To find $\sin{\theta}$ , just use opposite over hypotenuse with the right triangle $\triangle DCE$ . This is equal to $\frac69=\boxed{23}$ | D | 23 |
5d34571583a698d4f5ebcb679250324f | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_13 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$ | Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
\[(8-L)(p+h)=50\]
\[(6.2-L)h=24\]
\[(11.2-L)p=26\]
With three equations and three variables, we need to find the value of $L$ .
Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$ . Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$ , so we get $h=\frac{16}{9}p$ .
Plugging into the second equation:
\[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\]
We can then subtract this from the third equation:
\[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]
Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{48}$ | D | 48 |
5d34571583a698d4f5ebcb679250324f | https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_13 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$ | Because Paula worked from \[8:00 \text{A.M.}\] to \[7:12 \text{P.M.}\] , she worked for 11 hours and 12 minutes = 672 minutes. Since there is $100-50-24=26$ % of the house left, we get the equation $26a=672$ . Because $672$ is $22$ mod $26$ , looking at our answer choices, the only answer that is $22$ $\text{mod}$ $26$ is $48$ . So the answer is $\boxed{48}$ | D | 48 |
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