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3dc50ddfbe989a9fc21f84fad0a2f600 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_6 | Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits
of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a
two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9$ | Denote $a$ and $b$ as the tens and units digit of $N$ , respectively. Then $N = 10a+b$ . It follows that $10a+b=ab+a+b$ , which implies that $9a=ab$ . Since $a\neq0$ $b=9$ . So the units digit of $N$ is $\boxed{9}$ | E | 9 |
401a2b02f30d7018adf389aeeb3a0b16 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_7 | When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?
$\textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$ | If $x$ is the number, then moving the decimal point four places to the right is the same as multiplying $x$ by $10000$ . This gives us: \[10000x=4\cdot\frac{1}{x} \implies x^2=\frac{4}{10000}\] Since \[x>0\implies x=\frac{2}{100}=\boxed{0.02}\] | C | 0.02 |
401a2b02f30d7018adf389aeeb3a0b16 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_7 | When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?
$\textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$ | Alternatively, we could try each solution and see if it fits the problems given statements.
After testing, we find that $\boxed{0.02}$ is the correct answer. | C | 0.02 |
6caf14171e1bf355d84a22a11e81b232 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_8 | Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today will they next be together tutoring in the lab?
$\textbf{(A) }42\qquad\textbf{(B) }84\qquad\textbf{(C) }126\qquad\textbf{(D) }178\qquad\textbf{(E) }252$ | We need to find the least common multiple of the four numbers given.
$\textrm{LCM}(3, 4, 6, 7) = \textrm{LCM}(3, 2^2, 2 \cdot 3, 7) = 2^2 \cdot 3 \cdot 7 = 84$
So the answer is $\boxed{84}$ | B | 84 |
227cbd9a094a48b5f95a026123843b8c | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_3 | problem_id
227cbd9a094a48b5f95a026123843b8c The state income tax where Kristin lives is le...
227cbd9a094a48b5f95a026123843b8c The state income tax where Kristin lives is le...
Name: Text, dtype: object | Let $A$ $T$ be Kristin's annual income and the income tax total, respectively. Notice that \begin{align*} T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\ &= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\ &= p\%\cdot A + 2\%\cdot(A - 28000) \end{align*} We are also given that \[T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A\] Thus, \[p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A\] \[2\%\cdot(A - 28000) = 0.25\%\cdot A\] Solve for $A$ to obtain $A = \boxed{32000}$ | B | 32000 |
7ff4ba16afc36e5a186da8cde4e10530 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_10 | If $x$ $y$ , and $z$ are positive with $xy = 24$ $xz = 48$ , and $yz = 72$ , then $x + y + z$ is
$\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$ | The first two equations in the problem are $xy=24$ and $xz=48$ . Since $xyz \ne 0$ , we have $\frac{xy}{xz}=\frac{24}{48} \implies 2y=z$ . We can substitute $z = 2y$ into the third equation $yz = 72$ to obtain $2y^2=72 \implies y=6$ and $2y=z=12$ . We replace $y$ into the first equation to obtain $x=4$
Since we know every variable's value, we can substitute them in to find $x+y+z = 4+6+12 = \boxed{22}$ | D | 22 |
7ff4ba16afc36e5a186da8cde4e10530 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_10 | If $x$ $y$ , and $z$ are positive with $xy = 24$ $xz = 48$ , and $yz = 72$ , then $x + y + z$ is
$\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf{(E) }24$ | These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives $(xyz)^2 = (xy)(yz)(xz) = (24)(48)(72) = (24 \times 12)^2 \implies xyz = 288$ . We divide $xyz = 288$ by each of the given equations, which yields $x = 4$ $y = 6$ , and $z = 12$ . The desired sum is $4+6+12 = 22$ , so the answer is $\boxed{22}$ | D | 22 |
a4fa095c5424a342f00ce4e9140f333a | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_11 | Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^\text{th}$ ring is
[asy] unitsize(3mm); defaultpen(linewidth(1pt)); fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black); for(real i=0; i<=9; ++i) { draw((i,0)--(i,9)); draw((0,i)--(9,i)); }[/asy]
$\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$ | We can partition the $n^\text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.
There are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^\text{th}$ ring.
Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{800}$ unit squares. | C | 800 |
a4fa095c5424a342f00ce4e9140f333a | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_11 | Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^\text{th}$ ring is
[asy] unitsize(3mm); defaultpen(linewidth(1pt)); fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black); for(real i=0; i<=9; ++i) { draw((i,0)--(i,9)); draw((0,i)--(9,i)); }[/asy]
$\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$ | We can make the $n^\text{th}$ ring by removing a square of side length $2n-1$ from a square of side length $2n+1$
This ring contains $(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$ unit squares.
Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{800}$ unit squares. | C | 800 |
a4fa095c5424a342f00ce4e9140f333a | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_11 | Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^\text{th}$ ring is
[asy] unitsize(3mm); defaultpen(linewidth(1pt)); fill((2,2)--(2,7)--(7,7)--(7,2)--cycle, mediumgray); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle, gray); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle, black); for(real i=0; i<=9; ++i) { draw((i,0)--(i,9)); draw((0,i)--(9,i)); }[/asy]
$\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$ | Notice that the first ring around the center square contains $8$ unit squares, the second ring contains $16$ unit squares, the third contains $24$ unit squares, and so on. The number of squares in the $n^\text{th}$ ring is determined by the expression $8 \times n$ . Thus, the number of unit squares in the $100^\text{th}$ ring is equal to $8 \times 100$ , which equals $\boxed{800}$ unit squares. | C | 800 |
cae5e063a616f7a3e0443c6e3f5813cc | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_12 | Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$ . Which of the following is not necessarily a divisor of $n$
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42$ | Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$ , meaning it is divisible by $6$
It also mentions that it is divisible by $7$ , so the number is definitely divisible by all the factors of $42$
In our answer choices, the one that is not a factor of $42$ is $\boxed{28}$ | D | 28 |
cae5e063a616f7a3e0443c6e3f5813cc | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_12 | Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$ . Which of the following is not necessarily a divisor of $n$
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42$ | We can look for counterexamples. For example, letting $n = 13 \cdot 14 \cdot 15$ , we see that $n$ is not divisible by 28, so $\boxed{28}$ is our answer. | D | 28 |
74b973e4f94621e9337c1a9c0077ccfc | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_6 | problem_id
74b973e4f94621e9337c1a9c0077ccfc A telephone number has the form $\text{ABC-DEF...
74b973e4f94621e9337c1a9c0077ccfc A telephone number has the form $\text{ABC-DEF...
Name: Text, dtype: object | We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
Case 1: $G$ $H$ $I$ , and $J$ are $7$ $5$ $3$ , and $1$ respectively.
A cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$ $E$ , and $F$ are $8$ $6$ , and $4$ respectively, so this is out of the question.
Case 2: $G$ $H$ $I$ , and $J$ are $3$ $5$ $7$ , and $9$ respectively.
A cursory glance allows us to deduce the answer. Clearly, when $D$ $E$ , and $F$ are $6$ $4$ , and $2$ respectively, $A + B + C$ is $9$ when $A$ $B$ , and $C$ are $8$ $1$ , and $0$ respectively, giving us a final answer of $\boxed{8}$ | E | 8 |
afa106734f55c02711ecd5e8bbf4e8d3 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.
Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.
Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single full price ticket costs $\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0\leq h\leq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.
The prime factorization of $4002$ is $4002=2\cdot 3\cdot 23\cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2\cdot 3\cdot 29 = 174$
This gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.
In our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $\frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23\cdot 34 = \boxed{782}$ dollars are raised by the full price tickets. | A | 782 |
afa106734f55c02711ecd5e8bbf4e8d3 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | Let the cost of the full price ticket be $x$ , the number of full-price tickets be $A$ , and the number of half-price tickets be $B$
Let's multiply both sides of the equation that naturally follows by 2. We have
\[2Ax+Bx=4002\]
And we have $A+B=140\implies B=140-A$
Plugging in, we get $\implies 2Ax+(140-A)(x)=4002$
Simplifying, we get $Ax+140x=4002$
Factoring out the $x$ , we get $x(A+140)=4002\implies x=\frac{4002}{A+140}$
We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)
Thus, $A+140$ must be a factor of 4002.
Consider the prime factorization of $4002$ $2\times3\times23\times29$
$A$ must be a positive integer. So, we seek a factor of $4002$ to set equal to $A+140$ so that we get an integer solution for $A$ that is less than $140$ . By guess-and-check OR inspection, the appropriate factor is $174$ $2\times3\times29$ ), meaning that $A$ has a value of $34$ . Plug this into the above equation for $x$ to get $x = 23$
Therefore, the price of full tickets out of $2001$ is $23\times34=\boxed{782}$ | A | 782 |
afa106734f55c02711ecd5e8bbf4e8d3 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | Let $f$ equal the number of full-price tickets, and let $h$ equal the number of half-price tickets. Additionally, suppose that the price of $f$ is $p$ . We are trying to solve for $f \cdot p$
Since the total number of tickets sold is $140$ , we know that \[f+h=140.\] The sales from full-price tickets ( $f \cdot p$ ) plus the sales from half-price tickets $\Big(\frac{h \cdot p}{2}$ , because each hall-price ticket costs $\frac{p}{2}$ dollars $\Big)$ equals $2001.$ Then we can write \[fx + \frac{hx}{2}=2001.\]
Substituting $h=140-f$ into the second equation, we get \[f \cdot p +\frac{(140-f)p}{2}=f \cdot p+\frac{140p-f\cdot p}{2}=\frac{f\cdot p+140p}{2}=2001.\]
Multiplying by $2$ and subtracting $140p$ gives us \[f\cdot p=4002-140p.\]
Since the problem states that $x$ is a whole number, $140p$ will be some integer multiple of $140$ that ends in a $0$ . Thus, $4002-140p$ will end in a $2$ . Looking at the answer choices, only $\boxed{782}$ satisfies that condition. | A | 782 |
7b5ba126e608735f52ecfa666d6832a4 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_15 | A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$ | Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$ , giving an area of $600$
If we look at it the other way, we see the distance between the stripes is the height and the base is $50$
[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed);[/asy]
The area is still the same, so the distance between the stripes is $600/50 = \boxed{12}$ | C | 12 |
7b5ba126e608735f52ecfa666d6832a4 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_15 | A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$ | Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $\boxed{12}$ | C | 12 |
ae79010feec50f73241383732e6c476e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | problem_id
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
Name: Text, dtype: object | Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, $5$ . So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which can be solved to get $m=10$ .
Hence, the sum of the three numbers is $3\cdot 10 = \boxed{30}$ | D | 30 |
ae79010feec50f73241383732e6c476e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | problem_id
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
Name: Text, dtype: object | Say the three numbers are $x$ $y$ and $z$ . When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$
We can also assume that the smallest number is $x$ and the largest number of the three is $y$ .
Therefore,
\[\frac{x+y+z}{3} = x + 10 = z - 15\] \[\frac{x+5+z}{3} = x + 10 = z - 15\]
Taking up the first equation $\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$
\[x - 2z + 50 = 2x - 4z + 100\]
when solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$
Therefore the sum of the three numbers is $25 + 5 + 0 = \boxed{30}$ | D | 30 |
c1c2900151c908ac390988a490c7e35c | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_10 | problem_id
c1c2900151c908ac390988a490c7e35c The plane is tiled by congruent squares and co...
c1c2900151c908ac390988a490c7e35c The plane is tiled by congruent squares and co...
Name: Text, dtype: object | Consider any single tile:
[asy] unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); [/asy]
If the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons.
Hence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.\overline{5}\%$ , and the closest integer to this value is $\boxed{56}$ | D | 56 |
30777bbe440e767547a25566a50f0fc5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_19 | Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18$ | Let's use stars and bars .
Let the donuts be represented by $O$ s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us $4$ in all. The four donuts we want can be represented as $OOOO$ . Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, $O|OO|O$ represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in $\binom{6}{2}=15$ ways. Our answer is hence $\boxed{15}$ . Notice that this can be generalized to get the stars and bars (balls and urns) identity. | D | 15 |
30777bbe440e767547a25566a50f0fc5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_19 | Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18$ | Simple casework works here as well:
Set up the following ratios: \[4:0:0\] \[3:1:0\] \[2:2:0\] \[2:1:1\]
In three of these cases we see that there are two of the same ratios (so like two boxes would have $0$ ), and so if we swapped those two donuts, we would have the same case. Thus we get $\frac{3!}{2!1!}$ for those $3$ (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal $\binom{4}{3}=6.$ Thus, our answer is $3 \cdot 3+6 = \boxed{15}$ | D | 15 |
a78bcb57a15495309f8df5cde44e1ccf | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_22 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ $w$ $x$ $y$ , and $z$ . Find $y + z$
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$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | We know that $y+z=2v$ , so we could find one variable rather than two.
$v+24+w=43+v$
$24+w=43$
$w=19$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]
$44+x=24+x+z \implies z=20$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]
The sum per row is $25+21+20=66$
Thus $66-18-25=66-43=v=23$
Since we needed $2v$ and we know $v=23$ $23 \times 2 = \boxed{46}$ | D | 46 |
a78bcb57a15495309f8df5cde44e1ccf | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_22 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ $w$ $x$ $y$ , and $z$ . Find $y + z$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | $v+24+w=43+v$
$24+w=43$
$w=19$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]
$44+x=24+x+z \implies z=20$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$20$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$19$",(2.5,2.5));[/asy]
The magic sum is determined by the bottom row. $25+20+21=66$
Solving for $y$
$y=66-19-21=66-40=26$
To find our answer, we need to find $y+z$ $y+z=20+26 = \boxed{46}$ | D | 46 |
a78bcb57a15495309f8df5cde44e1ccf | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_22 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ $w$ $x$ $y$ , and $z$ . Find $y + z$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | A nice thing to know is that any $3$ numbers that go through the middle form an arithmetic sequence.
Using this, we know that $x=(24+z)/2$ , or $2x=24+z$ because $x$ would be the average.
We also know that because $x$ is the average the magic sum would be $3x$ , so we can also write the equation $3x-46=z$ using the bottom row.
Solving for x in this system we get $x=22$ , so now using the arithmetic sequence knowledge we find that $y=26$ and $z=20$
Adding these we get $\boxed{46}$ | D | 46 |
a78bcb57a15495309f8df5cde44e1ccf | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_22 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$ $w$ $x$ $y$ , and $z$ . Find $y + z$
[asy] unitsize(10mm); defaultpen(linewidth(1pt)); for(int i=0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$25$",(0.5,0.5)); label("$z$",(1.5,0.5)); label("$21$",(2.5,0.5)); label("$18$",(0.5,1.5)); label("$x$",(1.5,1.5)); label("$y$",(2.5,1.5)); label("$v$",(0.5,2.5)); label("$24$",(1.5,2.5)); label("$w$",(2.5,2.5));[/asy]
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | Create an equation for every row, column, and diagonal. Let $e$ be the sum of the rows, columns, and diagonals. \[w+v+24=e\] \[x+y+18=e\] \[z+46=e\] \[v+43=e\] \[x+z+24=e\] \[w+y+21=e\] \[x+w+25=e\] \[x+v+21=e\]
Notice that $z+46=e$ and $x+z+24=e$ both have $z$ . Equate them and you get that $x=22$ .
Using that same strategy, we use $v+43=e$ instead. $w+v+24=e$ is good for our purposes. It turns out that $w=19$ . Since we already know those numbers, and $x+w+25=e$ , We can say that $e$ will be $66$ . We are now able to solve: $x+z+24=e$ $w+y+21=e$ $x+v+21=e$ , and $x+y+18=e$ . Respectively, $v=23$ $w=19$ $x=22$ $y=26$ , and $z=20$ . We only require The sum of $y+z$ , which is $26+20=46$ .
We get that the sum of $y$ and $z$ respectively is $\boxed{46}$ | D | 46 |
44dac98b900fb2d03612e3e20d26762f | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | problem_id
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
Name: Text, dtype: object | Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: $W, W, R, R, R$ such that both $W$ 's appear in the first $4$ . We find the number of ways to arrange the white chips in the first $4$ and divide that by the total ways to choose all the chips. The probability of this occurring is $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{35}$ | D | 35 |
44dac98b900fb2d03612e3e20d26762f | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | problem_id
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
Name: Text, dtype: object | The amount of ways to end with a white chip is by having $RRWW, RWW,$ and $WW$ . The amount of arrangements for $RRWW$ with $W$ at the end is $3$ , the number of arrangements of $RWW$ with $W$ at the end is $2$ , and the number of arrangements with $WW$ is just $1$ . This gives us $6$ total ways to end with white. Next, the cases to end with a red are $RWRR$ , and $RRR$ $RWRR$ gives us $3$ ways and $RRR$ gives us $1$ way. So the number of ways to end with a red is $4$ . Thus, our answer is simply $\frac{6}{4+6}$ $\boxed{35}$ | D | 35 |
8aef7612e54a2d764bbb5e8af5cf5d20 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_24 | In trapezoid $ABCD$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 7.3, ymin = -3.16, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((0.2,4.92), 1.3)); draw(circle((1.04,1.58), 2.14)); draw((-1.1,4.92)--(0.2,4.92)); draw((0.2,4.92)--(1.04,1.58)); draw((1.04,1.58)--(-1.1,1.58)); draw((-1.1,1.58)--(-1.1,4.92)); /* dots and labels */ dot((-1.1,4.92),dotstyle); label("$A$", (-1.02,5.12), NE * labelscalefactor); dot((0.2,4.92),dotstyle); label("$B$", (0.28,5.12), NE * labelscalefactor); dot((-1.1,1.58),dotstyle); label("$D$", (-1.02,1.78), NE * labelscalefactor); dot((1.04,1.58),dotstyle); label("$C$", (1.12,1.78), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
If $AB=x$ and $CD=y$ , then $BC=x+y$ . By the Pythagorean theorem , we have $(x+y)^2=(y-x)^2+49.$ Solving the equation, we get $4xy=49 \implies xy = \boxed{12.25}$ | B | 12.25 |
8aef7612e54a2d764bbb5e8af5cf5d20 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_24 | In trapezoid $ABCD$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | Simpler is just drawing the trapezoid and then using what is given to solve.
Draw a line parallel to $\overline{AD}$ that connects the longer side to the corner of the shorter side. Name the bottom part $x$ and top part $a$ .
By the Pythagorean theorem, it is obvious that $a^{2} + 49 = (2x+a)^{2}$ (the RHS is the fact the two sides added together equals that). Then, we get $a^2 + 49 = 4x^2 + 4ax + a^2$ , cancel out and factor and we get $49 = 4x(x+a)$ . Notice that $x(x+a)$ is what the question asks, so the answer is $\boxed{12.25}$ | B | 12.25 |
8aef7612e54a2d764bbb5e8af5cf5d20 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_24 | In trapezoid $ABCD$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | We know it is a trapezoid and that $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ . If they are perpendicular to $\overline{AD}$ that means this is a right-angle trapezoid (search it up if you don't know what it looks like or you can look at the trapezoid in the first solution). We know $\overline{AD}$ is $7$ . We can then set the length of $\overline{AB}$ to be $x$ and the length of $\overline{DC}$ to be $y$ $\overline{BC}$ would then be $x+y$ . Let's draw a straight line down from point $B$ which is perpendicular to $\overline{DC}$ and parallel to $\overline{AD}$ . Let's name this line $M$ . Then let's name the point at which line $M$ intersects $\overline{DC}$ point $E$ . Line $M$ partitions the trapezoid into ▭ $ADEB$ and $\triangle$ $BEC$ . We will use the triangle to solve for $xy$ using the Pythagorean theorem. The line segment $\overline{EC}$ would be $y-x$ because $\overline{DC}$ is $y$ and $\overline{DE}$ is $x$ $\overline{DE}$ is $x$ because it is parallel to $\overline{AB}$ and both are of equal length. Because of the Pythagorean theorem, we know that $(EC)^2+(BE)^2=(BC)^2$ . Substituting the values we have we get $(y-x)^2+(7)^2=(x+y)^2$ . Simplifying this we get $(y^2-2xy+x^2)+(49)=(x^2+2xy+y^2)$ . Now we get rid of the $x^2$ and $y^2$ terms from both sides to get $(-2xy)+(49)=(2xy)$ . Combining like terms we get $(49)=(4xy)$ . Then we divide by $4$ to get $(12.25)=(xy)$ . Now we know that $xy\ =\ 12.25$ which is answer choice $\boxed{12.25}$ | B | 12.25 |
8aef7612e54a2d764bbb5e8af5cf5d20 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_24 | In trapezoid $ABCD$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | Choose any value for $BC$ , and then use Pythagorean theorem to get $CD - AB$ , and $AB = (BC - (CD - AB))/2$ ). Then multiply $AB \cdot CD$
For example:
$BC=25$ $CD - AB = \sqrt{25^2 - 7^2} = 24$ $AB = (25 - 24)/2=0.5$ $CD = 0.5 + 24 = 24.5$ $AB \cdot CD = (0.5)(24.5)= \boxed{12.25}$ | B | 12.25 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\left\lfloor \frac{2001}{3} \right\rfloor$ $\left\lfloor \frac{2001}{4} \right\rfloor$ , and $\left\lfloor \frac{2001}{12} \right\rfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.
The second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\left\lfloor \frac{2001}{3\cdot5} \right\rfloor$ $\left\lfloor \frac{2001}{4\cdot5} \right\rfloor$ , and $\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\rfloor$ yields the numbers $133$ $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$
There are $\frac45\cdot2000=1600$ numbers that are not multiples of $5$ $\frac23\cdot\frac34\cdot1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$ , so there are $\frac{10}{25}\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$ . Since $2001$ is a multiple of $3$ , add $1$ to $800$ to get $800+1=\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$ $80\%$ of them will not be divisible by $5$ , and by far the closest number to $80\%$ of $1001$ is $\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | Similar to some of the above solutions.
We can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$ . Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$ , and $500$ numbers that are divisible by $4$ . However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$ , since they're also divisible by 5 which we don't want. There are $133$ $100$ $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$ , so we have to add $33$ back to the tally (there are $33$ multiples of $60$ that does not exceed $2001$ ). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$ . This is tantamount to subtracting the number of multiples of $12$ . And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ $500$ $133$ $100$ $33$ $166$ $\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | Similar to @above:
Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$ . Then we have that the desired number is: \[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)\]
Evaluating each of these we get: \[667+500-166-133-100+33 = 1100-299 = 801.\]
Thus, the answer is $\boxed{801}.$ | B | 801 |
59ed238522fce629e7eff4df3baa8470 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_1 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$ | First, we need to recognize that a number is going to be lowest only if, of the $3$ factors , two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$ . It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$ . Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors.
Hence, we have $667 + 3 + 1 = \boxed{671}$ | E | 671 |
59ed238522fce629e7eff4df3baa8470 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_1 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$ | The sum is the highest if two factors are the lowest.
So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{671}$ | E | 671 |
59ed238522fce629e7eff4df3baa8470 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_1 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$ | We see since $2 + 0 + 0 + 1$ is divisible by $3$ , we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{671}$ as our final answer. | E | 671 |
b655d8c19fdcd7df044038b51a9a6109 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_3 | Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$ | We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$
After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \boxed{50}$ | B | 50 |
b655d8c19fdcd7df044038b51a9a6109 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_3 | Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$ | Testing the answers choices out, we see that the answer is $\boxed{50}$ | B | 50 |
841f24604dfa5cefb09df7827e69a3a2 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_5 | Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?
(a) the length of the segment $MN$
(b) the perimeter of $\triangle PAB$
(c) the area of $\triangle PAB$
(d) the area of trapezoid $ABNM$
[asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N); [/asy]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | (a) Triangles $ABP$ and $MNP$ are similar, and since $PM=\frac{1}{2}AP$ $MN=\frac{1}{2}AB$
(b) We see the perimeter changes. For example, imagine if P was extremely far to the left.
(c) The area clearly doesn't change, as both the base $AB$ and its corresponding height remain the same.
(d) The bases $AB$ and $MN$ do not change, and neither does the height, so the area of the trapezoid remains the same.
Only $1$ quantity changes, so the correct answer is $\boxed{1}$ | B | 1 |
70a31fe42a64a93c44a3d4d0462a8f8c | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_6 | The Fibonacci sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
$\textbf{(A)} \ 0 \qquad \textbf{(B)} \ 4 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 9$ | Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$
$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$
The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \boxed{6}$ | C | 6 |
c27cfbf7071135517b6059565b96c7bd | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_10 | The sides of a triangle with positive area have lengths $4$ $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10$ | Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$ . This gives us the triangle inequality $2<x<10$ and $2<y<10$ $7$ can be attained by letting $x=9.1$ and $y=2.1$ . However, $8=10-2$ cannot be attained. Thus, the answer is $\boxed{8}$ | D | 8 |
f40840caf9c4575ebd88f99971e07a90 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_11 | Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$ | Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $\boxed{119}$ | C | 119 |
f40840caf9c4575ebd88f99971e07a90 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_11 | Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$ | Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick , we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $\boxed{119}$ | C | 119 |
bf5899e1988e4eb98ddb2495b4226041 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_12 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ . Thus, the original sequence can be generated from a quadratic function.
If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables:
$f(0) = 1$ gives $c = 1$
$f(1) = 5$ gives $a + b + c = 5$
$f(2) = 13$ gives $4a + 2b + c = 13$
Plugging in $c=1$ into the last two equations gives
$a + b = 4$
$4a + 2b = 12$
Dividing the second equation by 2 gives the system:
$a + b = 4$
$2a + b = 6$
Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ . Thus, our quadratic function is:
$f(n) = 2n^2 + 2n + 1$
Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $\boxed{20201}$ | C | 20201 |
bf5899e1988e4eb98ddb2495b4226041 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_12 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{n=1}^{100} n = 5050$ squares. $4 \cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = \boxed{20201}$ | C | 20201 |
bf5899e1988e4eb98ddb2495b4226041 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_12 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | Let $a_n$ be the number of squares in figure $n$ . We can easily see that \[a_0=4\cdot 0+1\] \[a_1=4\cdot 1+1\] \[a_2=4\cdot 3+1\] \[a_3=4\cdot 6+1.\] See that we multiply the number we are on to the next consecutive number.
Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number. Hence, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{20201}$ .
~ edited by mathlover66 | C | 20201 |
bf5899e1988e4eb98ddb2495b4226041 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_12 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | Let $f_n$ denote the number of unit cubes in a figure. We have \[f_0=1\] \[f_1=5\] \[f_2=13\] \[f_3=25\] \[f_4=41\] \[...\]
Computing the difference between the number of cubes in each figure yields \[4,8,12,16,...\] It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$
From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields
\[S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}\] \[=50(2\cdot 4+99\cdot 4)\] \[=50(101\cdot 4)\] \[=200\cdot 101\] \[=20200\]
So $f_{100}=1+20200=\boxed{20201}$ unit cubes. | C | 20201 |
bf5899e1988e4eb98ddb2495b4226041 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_12 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$ | Newton’s little formula states that $a_n = A \binom{n-1}{0} + B \binom{n-1}{1} + C \binom{n-1}{2} + \cdots + K \binom{n-1}{m}$ if first term is $a_1$ and $A =$ first difference, $B =$ second difference, and so on. Hence we apply the formula (because we start at term 0, term 100 is $a_{101}$ ): $a_{101} = 5 \binom{100}{0} + 8 \binom{100}{1} + 4 \binom{100}{2} = \boxed{20201}.$ ~Peelybonehead
~clarification by LeonidasTheConquerer | C | 20201 |
faa3845fdebe850cd18a2194208e71c1 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_20 | Let $A$ $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
$\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89$ | The trick is to realize that the sum $AMC+AM+MC+CA$ is similar to the product $(A+1)(M+1)(C+1)$ . If we multiply $(A+1)(M+1)(C+1)$ , we get \[(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.\] We know that $A+M+C=10$ , therefore $(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11$ and \[AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.\] Now consider the maximal value of this expression. Suppose that some two of $A$ $M$ , and $C$ differ by at least $2$ . Then this triple $(A,M,C)$ is not optimal. (To see this, WLOG let $A\geq C+2.$ We can then increase the value of $(A+1)(M+1)(C+1)$ by changing $A \to A-1$ and $C \to C+1$ .)
Therefore the maximum is achieved when $(A,M,C)$ is a rotation of $(3,3,4)$ . The value of $(A+1)(M+1)(C+1)$ in this case is $4\cdot 4\cdot 5=80,$ and thus the maximum of $AMC + AM + MC + CA$ is $80-11 = \boxed{69}.$ | C | 69 |
faa3845fdebe850cd18a2194208e71c1 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_20 | Let $A$ $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
$\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89$ | Notice that if we want to maximize $AMC + AM + MC + AC$ , we want A, M, and C to be as close as possible. For example, if $A = 7, B = 2,$ and $C=1,$ then the expression would have a much smaller value than if we were to substitute $A = 4, B = 5$ , and $C = 1$ . So to make A, B, and C as close together as possible, we divide $\frac{10}{3}$ to get $3$ . Therefore, A must be 3, M must be 3, and C must be 4. $AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69$ . So the answer is $\boxed{69.}$ | C | 69. |
faa3845fdebe850cd18a2194208e71c1 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_20 | Let $A$ $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
$\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89$ | According to our knowledge in AM-GM, the closer numbers are, the value of their product is bigger. Assume that $A=B<C$ , we can get the set $A=3,B=3,C=4$ which the answer is $\boxed{69.}$ ~bluesoul | C | 69. |
87575dca7aa1a4a0058aff57fc980bae | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_22 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7$ | If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\boxed{5}$ | null | 5 |
87575dca7aa1a4a0058aff57fc980bae | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_22 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7$ | The number of ounces that Angela's family drank has to be a multiple of 8, so we can find the right answer by guessing random values for the number of ounces of coffee and milk Angela drank. With Angela drinking 4 ounces of milk and 4 ounces of coffee, we get 40 total ounces Angela's family drank. Dividing that by 8(each person drank 8 ounces) we get 5 members who drank the coffee with milk, or $\boxed{5}$ ~ Mathyguy88 | C | 5 |
fd878ccb8104f1f178ddf35c4fa4941d | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_24 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Let $y = \frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\frac{1}{3}, \frac{2}{9}$ . These sum up to $\boxed{19}$ | B | 19 |
fd878ccb8104f1f178ddf35c4fa4941d | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_24 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | This is quite trivially solved, as $3x = \dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has solutions $-\frac{1}{3}$ and $\frac{2}{9}$ . Adding these yields a solution of $\boxed{19}$ | B | 19 |
fd878ccb8104f1f178ddf35c4fa4941d | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_24 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{19}$ | B | 19 |
fd878ccb8104f1f178ddf35c4fa4941d | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_24 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-\frac{1}{9}$ or $\boxed{19}$ | B | 19 |
fd878ccb8104f1f178ddf35c4fa4941d | https://artofproblemsolving.com/wiki/index.php/2000_AMC_10_Problems/Problem_24 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Since we have $f(x/3)$ $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7:
$81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-\frac{b}{a}$ . Thus, the sum of the roots of this equation is $-\frac{9}{81} = \boxed{19}$ | B | 19 |
69755f5c1c2aab0c678b671d5ed1edd5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$ | E | 27 |
69755f5c1c2aab0c678b671d5ed1edd5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$ | E | 27 |
69755f5c1c2aab0c678b671d5ed1edd5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\] | E | 27 |
69755f5c1c2aab0c678b671d5ed1edd5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbara) will meet in 1 1/2 hours at 18mph x 3/2 hours = 27 miles from A. $\boxed{27}$ | E | 27 |
69755f5c1c2aab0c678b671d5ed1edd5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$ | E | 27 |
69755f5c1c2aab0c678b671d5ed1edd5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$ | Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total distance. | E | 27 |
e079894416dc23df8a285a2dd4276976 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$ | A | 8 |
e079894416dc23df8a285a2dd4276976 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$ , there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{8}$ perfect squares less than 2023. | A | 8 |
e079894416dc23df8a285a2dd4276976 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Since $5$ is prime, each solution must be divisible by $5^2=25$ . We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{8}$ positive perfect squares no greater than $80$ | A | 8 |
e079894416dc23df8a285a2dd4276976 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{8}$ solutions. | A | 8 |
19d4b904b168846f2b59b549ca83bdf9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_4 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$
$10^{15}$ gives us $15$ digits and $243$ gives us $3$ digits. $15+3=\text{\boxed{18}$ | E | 18 |
19d4b904b168846f2b59b549ca83bdf9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_4 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Multiplying it out, we get that $8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000$ . Counting, we have the answer is $\text{\boxed{18}$ ~andliu766 | E | 18 |
38abd2523b95190be4c0c2feed7a2263 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_7 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$ | Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{9}$ .
For curious readers, the numbers (in chronological order) are: | E | 9 |
38abd2523b95190be4c0c2feed7a2263 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_7 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$ | There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.
If $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.
If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.
Summing through all cases, there are $3 + 2 + 4 = \boxed{9}$ dates. | E | 9 |
38abd2523b95190be4c0c2feed7a2263 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_7 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$ | We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\boxed{9}$ solutions. | E | 9 |
052f45c5e4652078947cacaf8feb7b35 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_8 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.
We can write the following equations:
\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]
Multiplying $(x+1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]
Multiplying $(2)$ by $(a+3)$ and solving, we get: \[ax+33=ax+2a+3x+6\] \[33=2a+3x+6\] \[2a+3x=27\qquad (4)\]
Solving the system of equations for $(3)$ and $(4)$ , we find that $a=3$ and $x=\boxed{7}$ | D | 7 |
052f45c5e4652078947cacaf8feb7b35 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_8 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Suppose Maureen took $n$ tests with an average of $m$
If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$
Cross-multiplying: $nm+11=nm+n+m+1$ , so $n+m=10$
If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$
Cross-multiplying: $nm+33=nm+2n+3m+6$ , so $2n+3m=27$
But $2n+3m$ can also be written as $2(n+m)+m=20+m$ . Therefore $m=27-20=\boxed{7}$ | D | 7 |
052f45c5e4652078947cacaf8feb7b35 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_8 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.
So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$
We can use the first equation to write $s$ in terms of $t$
We then substitute this into the second equation: $\frac{-t^2+10t+33}{t+3} = \frac{-t^2+10}{t}+2$
From here, we solve for t, getting $t=3$
We substitute this to get $s=21$
Therefore, the solution to the problem is $\frac{21}{3}=$ $\boxed{7}$ | D | 7 |
052f45c5e4652078947cacaf8feb7b35 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_8 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }6\qquad\textbf{(D) }7\qquad\textbf{(E) }8$ | Let's consider all the answer choices. If the average is $8$ , then, we can assume that all her test choices were $8$ . We can see that she must have gotten $8$ twice, in order for another score of $11$ to bring her average up by one. However, adding three $11$ 's will not bring her score up to 10. Continuing this process for the answer choices, we see that the answer is $\boxed{7}$ ~andliu766 | D | 7 |
7389b4de6b4f373153da1e20b1c9ea63 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | Because $y^3=x^2$ , set $x=a^3$ $y=a^2$ $a\neq 0$ ). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$ . Solve the equation to get $a=3$ or $-1$ . Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{36}$ | D | 36 |
7389b4de6b4f373153da1e20b1c9ea63 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$ . Solving the case where $y-x=+2y$ , we'd find that $x=-y$ . This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$ . This means that $x=3y$ . Inputting this value into the first equation, we find: \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$ | null | 36 |
7389b4de6b4f373153da1e20b1c9ea63 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | first expand
$(y-x)^2 = 4y^2$
$y^2-2xy+x^2 = 4y^2$
$y^2-2yx+x^2 = 4y^2$
$x^2-2xy-3y^2 = 0$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
consider
a=1
b=-2y
c=-3y^2
$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$
$x=\frac{2y\pm\sqrt{16y^2}}{2a}$
$\frac{2y+4y^{2}}{2}$ or $\frac{2y-4y^{2}}{2}$
$x=y+2y$ and $x=y-2y$
$x=3y$ and $x=-y$
we can see both x and y will be postive in $x=3y$
now do same as solution 2 \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$ | null | 36 |
7389b4de6b4f373153da1e20b1c9ea63 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | Since $a^2 = |a|^2$ , we can rewrite the second equation as $(x-y)^2=4y^2$
Let $u=x+y$ . The second equation becomes
\[(u-2y)^2 = 4y^2\] \[u^2 - 4uy = 0\] \[u = 4y\] \[x+y = 4y\] \[x = 3y.\]
Substituting this into the first equation, we have
\[y^3 = (3y)^2,\] so $x = 9$
Hence $x = 27$ and $x + y = \boxed{36}.$ | D | 36 |
7389b4de6b4f373153da1e20b1c9ea63 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$ | We will use the difference of squares in the second equation.
\[(y-x)^2=4y^2\] \[(y-x)^2-(2y)^2=0\] \[(y-x-2y)(y-x+2y)=0\] \[-(x+y)(3y-x)=0\]
Since x and y are positive, x+y is non-zero. Thus, \[3y=x\]
Substituting into the first equation:
\[y^3=x^2\] \[y^3=9y^2\] \[y=9, x=27 \rightarrow x+y=\boxed{36}\] | D | 36 |
c2153c3f388451a20a6bfff2c8fe84d4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ | Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$ -axis. $k_1=2=\tan \alpha$ $k_2=\dfrac{1}{3}=\tan \beta$ . The angle formed by the two lines is $\alpha-\beta$ $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$ . Therefore, $\alpha-\beta=\boxed{45}$ | C | 45 |
c2153c3f388451a20a6bfff2c8fe84d4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ | We can take any two lines of this form, since the angle between them will always be the same. Let's take $y=2x$ for the line with slope of 2 and $y=\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$ $(1,2)$ , and $(3,1)$ . The distance between the origin and $(1,2)$ is $\sqrt{5}$ . The distance between the origin and $(3,1)$ is $\sqrt{10}$ . The distance between $(1,2)$ and $(3,1)$ is $\sqrt{5}$ . We notice that we have a triangle with 3 side lengths: $\sqrt{5}$ $\sqrt{5}$ , and $\sqrt{10}$ . This forms a 45-45-90 triangle, meaning that the angle is $\boxed{45}$ | null | 45 |
c2153c3f388451a20a6bfff2c8fe84d4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ | Follow Solution 2 up until the lattice points section. Let's use $(0,0)$ $(2,4)$ , and $(9,3)$ . The distance between the origin and $(2,4)$ is $\sqrt{20}$ . The distance between the origin and $(9,3)$ is $\sqrt{90}$ . The distance between $(2,4)$ and $(9,3)$ is $\sqrt{50}$ . Using the Law of Cosines, we see the $50 = 90 + 20 - 2\times\sqrt{20}$ $\times\sqrt{90}$ $\times\cos(\theta)$ , where $\theta$ is the angle we are looking for.
Simplifying, we get $-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)$
$30 = \sqrt{1800} \times\cos(\theta)$
$30 = 30\sqrt{2} \times\cos(\theta)$
$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)$
Thus, $\theta = \boxed{45}$ | C | 45 |
c2153c3f388451a20a6bfff2c8fe84d4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ | We can set up vectors $\vec{a} = <1,2>$ and $\vec{b} = <3,1>$ to represent the two lines. We know that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \cos \theta$ . Plugging the vectors in gives us $\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$ . From this we get that $\theta = \boxed{45}$ | C | 45 |
c2153c3f388451a20a6bfff2c8fe84d4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ | Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ \begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\ 1+2i&=3r\cos\theta - r\sin\theta + 3ri\sin\theta + ri\cos\theta \\ \end{align*}
From this we have: \begin{align} 1 &= 3r\cos\theta - r\sin\theta \\ 2 &= r\cos\theta + 3r\sin\theta \end{align}
To solve this we must compute $r$ \begin{align*} r &= \frac{|Z_2|}{|Z_1|} \\ &= \frac{\sqrt{5}}{\sqrt{10}} \\ &= \frac{\sqrt{2}}{2} \end{align*}
Using elimination we have: $3\cdot(2) - (1)$ \begin{align*} 5 &= 10r\sin\theta \\ \frac{1}{2r} &= \sin\theta \\ \frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\ \frac{\sqrt{2}}{2} &= \sin\theta \\ \theta &= \boxed{45} | C | 45 |
c2153c3f388451a20a6bfff2c8fe84d4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$ | The lines $y = 2x, y = \frac {1}{3}x$ , and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\alpha$ , and call the angle that is formed by the x-axis and the line $y = \frac {1}{3}x$ $\beta$ . We try to find $\sin (\alpha - \beta)$ first, and then try to see if any of the answer choices match up.
$\sin (\alpha - \beta)$ $\sin \alpha$ $\cos \beta$ $\sin \beta$ $\cos \alpha$
Using soh-cah-toa, we find that $\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5},$ and $\cos \beta = \frac {3}{\sqrt 10}$
Plugging it all in, we find that $\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}$ , which is equivalent to $\frac {\sqrt 2}{2}$ . Since $\sin (\alpha - \beta) = \frac {\sqrt 2}{2}$ , we get that $\alpha - \beta = 45^{\circ}$ . Therefore, the answer is $\boxed{45}$ | C | 45 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.
\[2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3\]
$=(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)$
$=(2^2+1 \cdot 2+1^2)+(4^2+4 \cdot 3+3^2)+(6^2+6 \cdot 5+5^2)+...+(18^2+18 \cdot 17+17^2)$
$=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$
$=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$
we could rewrite the second part as $\sum_{n=1}^{9}(2n-1)(2n)$
$(2n-1)(2n)=4n^2-2n$
$\sum_{n=1}^{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$
$\sum_{n=1}^{9}-2n=-2(\frac{9(9+1)}{2})$
Hence,
$1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18 = 4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
Adding everything up:
$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$
$=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
$=3(19)(37)+6(10)(19)-9(10)$
$=2109+1140-90$
$=\boxed{3159}$ | D | 3159 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | Think about $2^3+4^3+6^3+...+18^3$ . Once we factor out $2^3=8$ , we get $1^3+2^3+...+9^3$ , which can be found using the sum of cubes formula, $(\frac{n(n+1)}{2})^2$ . Now think about $1^3+3^3+...+17^3$ . This is just the previous sum subtracted from the total sum of $18$ cubes. So now we have the two things we need to add. The sum of all the even cubes is $8\cdot (\frac{90}{2})^2\rightarrow 8\cdot 2025 = 16200$ . The sum of all cubes from $1^3$ to $18^3$ is $(\frac{18\cdot 19}{2})^2=29241$ . The sum of the odd cubes is then $29241-16200=13041$ . Thus we get $16200-13041=\boxed{3159}$ ~amcrunner | D | 3159 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | Using the same sum of cubes formula, we can rewrite as $2(2^3 + 4^3 + ... + 18^3) - (1^3 + 2^3 + ... + 18^3)$
$= 2(2^3)(1^3 + ... + 9^3) - (1^3 + ... + 18^3)$
$= 16(5 \cdot 9)^2 - (9 \cdot 19)^2 = 9^2(20^2 - 19^2) = 81 \cdot 39 = \boxed{3159}$ ~AoPSuser216 | D | 3159 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | For any real numbers $x$ and $y$ $x^3-y^3=(x-y)(x^2+xy+y^2)=(x-y)((x-y)^2+3xy)=(x-y)^3+3xy(x-y)$
When $x = y + 1$ , with the above formula, we will get $x^3-y^3=1^3+3xy=1 + 3xy$
Therefore,
$2^3 - 1^3 + 4^3 - 3^3 + \dots + 18^3 - 17^3$
$= (1 + 3\cdot 1\cdot 2) + (1 + 3\cdot 3\cdot 4) + \dots + (1 + 3\cdot 17\cdot 18)$
$= 9 + 3\cdot (1\cdot 2 + 3\cdot 4 + \dots + 17\cdot 18)$
$= 9 + 3 \cdot (2 + 12 + 30 + 56 + 90 + 132 + 182 + 240 + 306)$
$= 9 + 3 \cdot 1050$
$= \boxed{3159}$ | D | 3159 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | We rewrite the sum as
\[\sum_{k=1}^{9}(2k)^3-(2k-1)^3\] \[=\sum_{k=1}^{9} 12k^2 - 6k + 1\] \[=12\sum_{k=1}^{9}k^2 - 6\sum_{k=1}^{9}k + 9\] \[=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9\] \[=3420 - 270 + 9 = \boxed{3159}\] | D | 3159 |
71f523979ced286fe544b672952956a5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 | What is the value of \[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
$\textbf{(A) } 2023 \qquad\textbf{(B) } 2679 \qquad\textbf{(C) } 2941 \qquad\textbf{(D) } 3159 \qquad\textbf{(E) } 3235$ | We see $=12\cdot\frac{9\cdot 10\cdot 19}{6} -6\cdot \frac{9\cdot 10}{2} + 9 = 9\cdot (12\cdot\frac{10\cdot 19}{6} -6\cdot \frac{10}{2} + 1)$ which is clearly a multiple of 9. The only answer choice which is a multiple of 9 is $\boxed{3159}$ ~Ilaggo2432 | D | 3159 |
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