id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
4,501 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 4 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean
is $87$ and the median is $90$
Thus, the solution is... | 3 |
4,502 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_5 | 5 | problem_id
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839 On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object | The percentage who scored $100$ points is $100\%-(10\%+35\%+30\%)=100\%-75\%=25\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\%$ .The lower $10\%+35\%=45\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\%$ is $100$ points and the lower $45\%$ is $7... | 3 |
4,503 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_6 | 1 | The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$ | Let the two digits be $a$ and $b$ . Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$ , or $2a = 7b$ . This yields $a = 7$ and $b = 2$ because $a, b < 10$ . Then, $72 + 27 = \boxed{99}.$ | 99 |
4,504 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_6 | 2 | The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$ | We start like above. Let the two digits be $a$ and $b$ . Therefore, $5(a+b) = 10a+b-10b-a=9(a-b)$ . Since we are looking for $10a+b+10b+a=11(a+b)$ and we know that $a+b$ must be a multiple of $9$ , the only answer choice that works is $\boxed{99}.$ | 99 |
4,505 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 5 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | Let $a=1$ . Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$ . Our next set starting with $3$ is $\{3,4,5,6,7\}$ . Our average is $25\div 5=5$
Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{4}$ | 4 |
4,506 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 6 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | We are given that \[b=\frac{a+a+1+a+2+a+3+a+4}{5}\] \[\implies b =a+2\]
We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]
Thus, the answer is $\boxed{4}$ | 4 |
4,507 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 7 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | We know from experience that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\boxed{4}$ | 4 |
4,508 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_10 | 8 | problem_id
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc Five positive consecutive integers starting wi...
Name: Text, dtype: object | The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$ . The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{4}$ | 4 |
4,509 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 4 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.
Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.
Simplifying gives $7d+525=10d$... | 210 |
4,510 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 5 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.
Quickly checking, we know that neither choice $\textbf{(A)}$ or choice $\textbf{(B)}$ work, but $\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \text{ mph}$... | 210 |
4,511 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_15 | 6 | problem_id
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b David drives from his home to the airport to c...
Name: Text, dtype: object | Let the total distance be $d$ . Then $d=35(t+1)$ . Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\frac{1}{2})=50(t-\frac{3}{2})$ . Solving, $d=35t+35=50t-40$ $15t=75$ $t=5$ , and $d=35(5+1)=35\cdot 6=210 \Longrightarr... | 210 |
4,512 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_13 | 1 | A fancy bed and breakfast inn has $5$ rooms, each with a distinctive color-coded decor. One day $5$ friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than $2$ friends per room. In how many ways can the innkeeper assign the gu... | We can work in reverse by first determining the number of combinations in which there are more than $2$ friends in at least one room. There are three cases:
Case 1: Three friends are in one room. Since there are $5$ possible rooms in which this can occur, we are choosing three friends from the five, and the other two f... | 220 |
4,513 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | 1 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$ ) palindromes. So the $a$ s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum.
For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$ ) palindromes. So the $b$ s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum.... | 18 |
4,514 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | 2 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | Notice that $10001+ 99999 = 110000.$ In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is $110000.$ We have $9\cdot 10\cdot 10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$ , s... | 18 |
4,515 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | 3 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | As shown above, there are a total of $900$ five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by $900$ to get our sum. The expected value for the ten-thousands and the units digit is $\frac{1+2+3+\cdot... | 18 |
4,516 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | 4 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | First, allow $a$ to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its $\textit{complement}$ . If $\overline{abcba}$ (the line means a, b, and c are digits and $abcba\ne a\cdot b\cdot c\cdot b\cdot a$ ) is a palindrome, then its complement is $\overline{defed}$ where $d=9-a$ $e... | 18 |
4,517 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_20 | 2 | problem_id
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
6a2cc5541e1749d3d0bb739aacabadf4 The product $(8)(888\dots8)$ , where the secon...
Name: Text, dtype: object | We can list the first few numbers in the form $8 \cdot (8....8)$
(Hard problem to do without the multiplication, but you can see the pattern early on)
$8 \cdot 8 = 64$
$8 \cdot 88 = 704$
$8 \cdot 888 = 7104$
$8 \cdot 8888 = 71104$
$8 \cdot 88888 = 711104$
By now it's clear that the numbers will be in the form $7$ $k-2$... | 991 |
4,518 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | 1 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }... | For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that:
Therefore, we have \begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}... | 271 |
4,519 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | 2 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }... | For simplicity, let $a=\log_{\frac{1}{16}}{x},b=\log_{16}a,c=\log_{\frac{1}{4}}b$ , and $d=\log_4c$
The domain of $\log_{\frac{1}{2}}x$ is $x \in (0, \infty)$ , so $d \in (0, \infty)$ .
Thus, $\log_4{c} \in (0, \infty) \Rightarrow c \in (1, \infty)$ .
Since $c=\log_{\frac{1}{4}}b$ we have $b \in \left(0, \left(\frac{1}... | 271 |
4,520 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_18 | 3 | The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }... | The domain of $f(x)$ is the range of the inverse function $f^{-1}(x)=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$ . Now $f^{-1}(x)$ can be seen to be strictly decreasing, since $\left(\frac12\right)^x$ is decreasing, so $4^{\left(\frac12\right)^x}$ is decreasing, so $\left(\frac14\r... | 271 |
4,521 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | 1 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | Factor the quadratic into \[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\] where $-n$ is our integer solution. Then, \[k = \frac{12}{n} + 5n,\] which takes rational values between $-200$ and $200$ when $|n| \leq 39$ , excluding $n = 0$ . This leads to an answer of $2 \cdot 39 = \boxed{78}$ | 78 |
4,522 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_19 | 2 | There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ . What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$ | Solve for $k$ so \[k=-\frac{12}{x}-5x.\] Note that $x$ can be any integer in the range $[-39,0)\cup(0,39]$ so $k$ is rational with $\lvert k\rvert<200$ . Hence, there are $39+39=\boxed{78}.$ | 78 |
4,523 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | 1 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sq... | Let $C_1$ be the reflection of $C$ across $\overline{AB}$ , and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$ . Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$ . (Proving this is a simple application of the triangle inequality; for an example of a simpler case,... | 14 |
4,524 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | 2 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sq... | In $\triangle BAC$ , the three lines look like the Chinese character 又. Let $\triangle DEA$ $\triangle CDA$ , and $\triangle BEA$ have bases $DE$ $CD$ , and $BE$ respectively. Then, $\triangle DEA$ has the same side $DA$ as $\triangle CDA$ and the same side $EA$ as $\triangle BEA$ . Connect all three triangles with $\t... | 14 |
4,525 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20 | 3 | In $\triangle BAC$ $\angle BAC=40^\circ$ $AB=10$ , and $AC=6$ . Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sq... | [asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,Ep,Bp,Cp; A = (0,0); B = 10*dir(-110);C = 6*dir(-70); Bp = 10*dir(-30);Cp = 6*dir(-150); D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C); draw(A--B--C--A--Cp--Bp--A); draw(Cp--B); draw(C--Bp); draw(C--D); draw(B--Ep)... | 14 |
4,526 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_21 | 1 | For every real number $x$ , let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$ , and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?
$\... | Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$ . Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$ . In order for this to be less than or equal to $1$ , we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$ . Combining this with the fact that $\lfloor x\rfloor =k$ gives ... | 1 |
4,527 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10A_Problems/Problem_25 | 2 | problem_id
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
8dd58d71d622cdca7859918e42e278d8 The number $5^{867}$ is between $2^{2013}$ and...
Name: Text, dtype: object | The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$
There can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$
The first power of $2$ is between $5^n$ and $2 \cdot 5^n$
The second power of $2$ is between $2 \cdot 5^n$ and $4 \cd... | 279 |
4,528 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_23 | 1 | The fraction
\[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\]
where $n$ is the length of the period of the repeating decimal expansion. What is the sum $b_0+b_1+\cdots+b_{n-1}$
$\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad$ | $\frac{1}{99^2}\\\\ =\frac{1}{99} \cdot \frac{1}{99}\\\\ =\frac{0.\overline{01}}{99}\\\\ =0.\overline{00010203...9799}$
So, the answer is $0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot10\cdot\frac{9\cdot10}{2}-(9+8)$ or $\boxed{883}$ | 883 |
4,529 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_25 | 1 | The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$ . For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$
$\textbf{(A) }38\qquad \textbf{(B) }40\qquad \textbf{(C) }42\qquad \textbf{(D) }44\qquad \textbf{(E) }46\qquad$ | The axis of $P$ is inclined at an angle $\theta$ relative to the coordinate axis, where $\tan\theta = \tfrac 34$ . We rotate the coordinate axis by angle $\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\widetilde{x}, \widetilde{y})$ be the coordi... | 40 |
4,530 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_1 | 1 | Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah's coins worth?
$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 37\qquad\textbf{(D)}\ 39\qquad\textbf{(E)}\ 41$ | She has $p$ pennies and $n$ nickels, where $n + p = 13$ . If she had $n+1$ nickels then $n+1 = p$ , so $2n+ 1 = 13$ and $n=6$ . So she has 6 nickels and 7 pennies, which clearly have a value of $\boxed{37}$ cents. | 37 |
4,531 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_2 | 1 | Orvin went to the store with just enough money to buy $30$ balloons. When he arrived he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\frac{1}{3}$ off the regular price. What is the greatest number of balloons Orvin could buy?
$\textbf{(A)}\ 33\qquad... | If every balloon costs $n$ dollars, then Orvin has $30n$ dollars. For every balloon he buys for $n$ dollars, he can buy another for $\frac{2n}{3}$ dollars. This means it costs him $\frac{5n}{3}$ dollars to buy a bundle of $2$ balloons. With $30n$ dollars, he can buy $\frac{30n}{\frac{5n}{3}} = 18$ sets of two balloo... | 36 |
4,532 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_5 | 1 | Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$ , and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill... | Let the height of the panes equal $5x$ , and let the width of the panes equal $2x$ . Now notice that the total width of the borders equals $10$ , and the total height of the borders is $6$ . We have \[10 + 4(2x) = 6 + 2(5x)\] \[x = 2\] Now, the total side length of the window equals \[10+ 4(2x) = 10 + 16 = \boxed{26}... | 26 |
4,533 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_6 | 1 | Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume $\frac{3}{4}$ of their drinks, Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both ... | Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\frac{3}{2}x$ ounces. After both consume $\frac{3}{4}$ of their drinks, Ed and Ann have $\frac{x}{4}$ and $\frac{3x}{8}$ ounces of their drinks remaining. Ann gives away $\frac{x}{8} + 2$ ounces to Ed.
In the end, Ed drank everything ... | 40 |
4,534 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_7 | 1 | For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | We know that $n \le 30$ or else $30-n$ will be negative, resulting in a negative fraction. We also know that $n \ge 15$ or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values $n$ from $15$ to $30$ gives us integer values for $... | 7 |
4,535 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_7 | 2 | For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let $\frac{n}{30-n}=m$ , where $m \in \mathbb{N}$ . Solving for $n$ , we find that $n=\frac{30m}{m+1}$ . Because $m$ and $m+1$ are relatively prime, $m+1|30$ . Our answer is the number of proper divisors of $2^13^15^1$ , which is $(1+1)(1+1)(1+1)-1 = \boxed{7}$ | 7 |
4,536 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_8 | 1 | In the addition shown below $A$ $B$ $C$ , and $D$ are distinct digits. How many different values are possible for $D$
\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$ . We know that $A+B = D$ . Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two ... | 7 |
4,537 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_9 | 1 | Convex quadrilateral $ABCD$ has $AB=3$ $BC=4$ $CD=13$ $AD=12$ , and $\angle ABC=90^{\circ}$ , as shown. What is the area of the quadrilateral?
[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,... | Note that by the pythagorean theorem, $AC=5$ . Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle. The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to \[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{36}\] | 36 |
4,538 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_10 | 1 | Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$
$\t... | We know that the number of miles she drove is divisible by $5$ , so $a$ and $c$ must either be the equal or differ by $5$ . We can quickly conclude that the former is impossible, so $a$ and $c$ must be $5$ apart. Because we know that $c > a$ and $a + c \le 7$ and $a \ge 1$ , we find that the only possible values for $... | 37 |
4,539 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_10 | 2 | Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$
$\t... | Let the number of hours Danica drove be $k$ . Then we know that $100a + 10b + c + 55k$ $100c + 10b + a$ . Simplifying, we have $99c - 99a = 55k$ , or $9c - 9a = 5k$ . Thus, k is divisible by $9$ . Because $55 * 18 = 990$ $k$ must be $9$ , and therefore $c - a = 5$ . Because $a + b + c \leq{7}$ and $a \geq{1}$ $a = 1$ $... | 37 |
4,540 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_11 | 1 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$ | We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we w... | 35 |
4,541 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_12 | 1 | A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \ge b \ge c$ $b+c > a$ , and $a, b, c < 5$ . Now we enumerate the elements of $T$
$(4, 4, 4)$
$(4, 4, 3)$
$(4, 4, 2)$
$(4, 4, 1)$
$(4, 3, 3)$
$(4, 3, 2)$
$(3, 3, 3)$
$(3, 3, 2)$
$(3, 3, 1)$
$(3, 2, 2)$
$(2, 2, 2)$
$(2, 2, 1)$
$(1, 1, 1)$
It shoul... | 9 |
4,542 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_16 | 1 | Let $P$ be a cubic polynomial with $P(0) = k$ $P(1) = 2k$ , and $P(-1) = 3k$ . What is $P(2) + P(-2)$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k$ | Let $P(x) = Ax^3+Bx^2 + Cx+D$ . Plugging in $0$ for $x$ , we find $D=k$ , and plugging in $1$ and $-1$ for $x$ , we obtain the following equations: \[A+B+C+k=2k\] \[-A+B-C+k=3k\] Adding these two equations together, we get \[2B=3k\] If we plug in $2$ and $-2$ in for $x$ , we find that \[P(2)+P(-2) = 8A+4B+2C+k+(-8A+4... | 14 |
4,543 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | 1 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}... | Let $y = m(x - 20) + 14$ . Equating them:
$x^2 = mx - 20m + 14$
$x^2 - mx + 20m - 14 = 0$
For there to be no solutions, the discriminant must be less than zero:
$m^2 - 4(20m - 14) < 0$
$m^2 - 80m + 56 < 0$
So $m < 0$ for $r < m < s$ where $r$ and $s$ are the roots of $m^2 - 80m + 56 = 0$ and their sum by Vieta's formul... | 80 |
4,544 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | 2 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}... | The line will begin to intercept the parabola when its slope equals that of the parabola at the point of tangency. Taking the derivative of the equation of the parabola, we get that the slope equals $2x$ . Using the slope formula, we find that the slope of the tangent line to the parabola also equals $\frac{14-x^2}{2... | 80 |
4,545 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_17 | 3 | Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}... | The smaller solution is basically negligible in comparison with the solution with the larger slope. Try some values like of $x^2 = y$ where $x = 40$ => $y = 1600$ , and slope ~80. Trying a few values leads us to conclude the least possible value is around $80$ , so the answer is $\boxed{80}$ | 80 |
4,546 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_24 | 2 | problem_id
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
77e0fabb4f8f99b4274e980292ddff5d The numbers $1, 2, 3, 4, 5$ are to be arranged...
Name: Text, dtype: object | We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is ... | 2 |
4,547 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_20 | 1 | For how many positive integers $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$
$\textbf{(A) }10\qquad \textbf{(B) }18\qquad \textbf{(C) }19\qquad \textbf{(D) }20\qquad \textbf{(E) }\text{infinitely many}\qquad$ | The domain of the LHS implies that \[40<x<60\] Begin from the left hand side \[\log_{10}[(x-40)(60-x)]<2\] \[(x-40)(60-x)<100\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{18}$ integers. | 18 |
4,548 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_23 | 1 | The number $2017$ is prime. Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$ . What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$ | Note that $2014\equiv -3 \mod2017$ . We have for $k\ge1$ \[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\] \[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\] Therefore \[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\] This is simply an alternating ser... | 24 |
4,549 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_25 | 1 | Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$ | Rewrite $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ . Now let $a = \cos{2x}$ , and let $b = \cos{\left( \frac{2014\pi^2}{x} \right) }$ . We have: \[2a(a - b) = 2a^2 - 2\]
Therefore, \[ab = 1\]
Notice that either $a = 1$ and $b = 1$ or $a = -1$ and $b = -1$ . For the first case, $a = 1$ only when $x = k\pi$ and $k$ is an int... | 80 |
4,550 | https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_25 | 2 | Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$ | Rewriting $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ and transposing $2\cos{2x}$ from the LHS to the RHS, we get,
\[\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}\] \[\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RH... | 80 |
4,551 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_5 | 1 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ $105$ , Dorothy paid $ $125$ , and Sammy paid $ $175$ . In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20... | Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: \[105 + t = 125 + d.\] Rearranging, we have \[t-d = \boxed{20}.\] | 20 |
4,552 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_5 | 2 | Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $ $105$ , Dorothy paid $ $125$ , and Sammy paid $ $175$ . In order to share the costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars. What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20... | Add up the amounts that Tom, Dorothy, and Sammy paid to get $ $405$ , and divide by 3 to get $ $135$ , the amount that each should have paid.
Tom, having paid $ $105$ , owes Sammy $ $30$ , and Dorothy, having paid $ $125$ , owes Sammy $ $10$
Thus, $t - d = 30 - 10 = 20$ , which is $\boxed{20}$ | 20 |
4,553 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_6 | 1 | In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D... | (similar to Solution 1, however a slightly more obvious way)
Say that
x = # of 2-pt shots
y = # of 3-pt shots
Because the total number of shots is $30$ $x + y = 30$
However, Shenille was only successful on $20\%$ of the 3-pt shots, and $30\%$ of the 2-pt shots, so
$0.2x + 0.3y$ = #number of successful shots
For each su... | 18 |
4,554 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_7 | 1 | The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, \[S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.\] Suppose that $S_9 = 110$ and $S_7 = 42$ . What is $S_4$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\te... | $S_9 = 110$ $S_7 = 42$
$S_8 = S_9 - S_ 7 = 110 - 42 = 68$
$S_6 = S_8 - S_7 = 68 - 42 = 26$
$S_5 = S_7 - S_6 = 42 - 26 = 16$
$S_4 = S_6 - S_5 = 26 - 16 = 10$
Therefore, the answer is $\boxed{10}$ | 10 |
4,555 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | 1 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | $x+\tfrac{2}{x}= y+\tfrac{2}{y}$
Since $x\not=y$ , we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$
Cross multiply in either equation, giving us $xy=2$
$\boxed{2}$ | 2 |
4,556 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | 2 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | Let $A = x + \frac{2}{x} = y + \frac{2}{y}.$ Consider the equation \[u + \frac{2}{u} = A.\] Reorganizing, we see that $u$ satisfies \[u^2 - Au + 2 = 0.\] Notice that there can be at most two distinct values of $u$ which satisfy this equation, and $x$ and $y$ are two distinct possible values for $u.$ Therefore, $x$ and ... | 2 |
4,557 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_8 | 3 | Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | \[x + \frac{2}{x} = y + \frac{2}{y}.\]
Multiply both sides by xy to get
\[x^2y + 2y = y^2x +2x\]
Rearrange to get
\[x^2y - y^2x = 2x - 2y\]
Factor out $xy$ on the left side and $2$ on the right side to get
\[xy(x-y) = 2(x-y)\]
Divide by $(x-y)$ {You can do this since x and y are distinct} to get
$\boxed{2}$ ~ e__ (the ... | 2 |
4,558 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_9 | 1 | In $\triangle ABC$ $AB=AC=28$ and $BC=20$ . Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively. What is the perimeter of parallelogram $ADEF$
[asy] siz... | Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $\triangle ABC$ , the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$ , and are therefore also isosceles triangles.
It follows that $BD = DE$ . Thus, $AD + DE = AD + DB = AB = 28$
Since opposite side... | 56 |
4,559 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_9 | 2 | In $\triangle ABC$ $AB=AC=28$ and $BC=20$ . Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively. What is the perimeter of parallelogram $ADEF$
[asy] siz... | We can set point $F$ to be on point $C$ , and point $D$ to be on point $A$
This makes a degenerate parallelogram with sides of length $28$ and $0$ , so it has a perimeter of $28 + 28 = \boxed{56}$ | 56 |
4,560 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_10 | 1 | Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{... | Let us begin by working with the condition $0.\overline{ab} = 0.ababab\cdots,$ . Let $x = 0.ababab\cdots$ . So, $100x-x = ab \Rightarrow x = \frac{ab}{99}$ . In order for this fraction $x$ to be in the form $\frac{1}{n}$ $99$ must be a multiple of $ab$ . Hence the possibilities of $ab$ are $1,3,9,11,33,99$ . Checking e... | 143 |
4,561 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_10 | 2 | Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{... | Notice that we have $\frac{100}{n}= ab.\overline{ab}$
We can subtract $\frac{1}{n}=00.\overline{ab}$ to get \[\frac{99}{n}=ab\]
From this we determine $n$ must be a positive factor of $99$
The factors of $99$ are $1,3,9,11,33,$ and $99$
For $n=1,3,$ and $9$ however, they yield $ab=99,33$ and $11$ which doesn't satisfy ... | 143 |
4,562 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_13 | 1 | Let points $A = (0,0) , \ B = (1,2), \ C = (3,3),$ and $D = (4,0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left (\frac{p}{q}, \frac{r}{s} \right )$ , where these fractions are in lowest terms. What is $p + q + r + s$
$\textbf{(... | If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.
Pick's Theorem states that
$A$ $I$ $+$ $\frac{B}{2}$ $1$ , where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice po... | 58 |
4,563 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_13 | 2 | Let points $A = (0,0) , \ B = (1,2), \ C = (3,3),$ and $D = (4,0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left (\frac{p}{q}, \frac{r}{s} \right )$ , where these fractions are in lowest terms. What is $p + q + r + s$
$\textbf{(... | Let the point of intersection be $E$ , with coordinates $(x, y)$ . Then, $ABCD$ is cut into $ABCE$ and $AED$
Since the areas are equal, we can use Shoelace Theorem to find the area. This gives $3 + 3x - 3y = 4y$
The line going through $CD$ is $y = -3x + 12$ . Since $E$ is on $CD$ , we can substitute this in, giving $3 ... | 58 |
4,564 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_15 | 1 | Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
$\textbf{(A)} \ 96 \qqua... | We tackle the problem by sorting it by how many stores are involved in the transaction.
1) 2 stores are involved. There are $\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \cdot 2 = 12$ total arrangements.
2) 3 stores are involved. There are $... | 204 |
4,565 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_16 | 1 | $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the grea... | Let the number of rocks in $A$ be $a$ $B$ be $b$ $C$ be $c$ . The total weight of $A$ be $40a$ $B$ be $50b$ $C$ be $kc$
We can write the information given as, $\frac{40a + 50b}{a+b} = 43$ $\frac{40a + kc}{a+c} = 44$ $\frac{50b + kc}{b+c} = ?$
$40a + 50b = 43 a + 43 b$ $3a = 7b$
$40a + kc = 44a + 44 c$ $kc = 4a + 44c = ... | 59 |
4,566 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_16 | 2 | $A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the grea... | Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: \[\frac{40A_n+50B_n}{A_n+B_n}=43\] .
Cross-multiplying, we get: \[40A_n+50B_n=43A_n+43B_n \implies 3A_n=7B_n\] .
Let's say we hav... | 59 |
4,567 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | 1 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to ... | The first pirate takes $\frac{1}{12}$ of the $x$ coins, leaving $\frac{11}{12} x$
The second pirate takes $\frac{2}{12}$ of the remaining coins, leaving $\frac{10}{12}\cdot \frac{11}{12}*x$
Note that
$12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}$
$11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot ... | 925 |
4,568 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | 2 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to ... | The answer cannot be an even number. Here is why:
Consider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some $\frac{n}{12}$ . This means the highest power of 2 that divides the ... | 925 |
4,569 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | 3 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to ... | Let $x$ be the number of coins the $12$ th pirate takes. Then the number of coins the $k<12$ th pirate takes is $\frac{12}{1} \cdot \frac{12}{2} \cdots \frac{12}{12-k} x$ . For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors, obviously there are sufficient $2... | 925 |
4,570 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_18 | 1 | Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$ . The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to th... | It can be seen that the diameter of the eighth sphere is equal to the radius of the seventh sphere by drawing out a diagram of the insides of the seventh sphere. The radius of the seventh sphere is $2+1=3$ , the radius of the eight sphere is $\boxed{32}$ | 32 |
4,571 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | 1 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61... | [asy] //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); [/asy]... | 61 |
4,572 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | 2 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61... | Let $BX = q$ $CX = p$ , and $AC$ meet the circle at $Y$ and $Z$ , with $Y$ on $AC$ . Then $AZ = AY = 86$ . Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , and that $p>13$ by the triangle inequality on $\triangle ACX$ . Thus, we get that $BC = p+q = \boxed{61}$ | 61 |
4,573 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_19 | 3 | In $\bigtriangleup ABC$ $AB = 86$ , and $AC = 97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61... | Let $x$ represent $CX$ , and let $y$ represent $BX$ . Since the circle goes through $B$ and $X$ $AB = AX = 86$ .
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$ , dividing both sides of the... | 61 |
4,574 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_23 | 1 | $ABCD$ is a square of side length $\sqrt{3} + 1$ . Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$ . The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$ , sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$ , where $a$ $b$ , and $c$ are positive integers and ... | We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$ . It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dime... | 19 |
4,575 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_25 | 1 | Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \... | Suppose $f(z)=z^2+iz+1=c=a+bi$ . We look for $z$ with $\operatorname{Im}(z)>0$ such that $a,b$ are integers where $|a|, |b|\leq 10$
First, use the quadratic formula:
$z = \frac{1}{2} (-i \pm \sqrt{-1-4(1-c)}) = -\frac{i}{2} \pm \sqrt{ -\frac{5}{4} + c }$
Generally, consider the imaginary part of a radical of a complex ... | 399 |
4,576 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_25 | 2 | Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \... | We consider the function $f(z)$ as a mapping from the 2-D complex plane onto itself. We complete the square of $f(z)=z^2+iz+1=(z+\frac{i}{2})^2+\frac{5}{4}$
Now, we must decide the range of $f(z)$ based on the domain of $z$ $\operatorname{Im}(z)>0$ . To do this, we are interested in mapping the boundary line $\operator... | 399 |
4,577 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_1 | 1 | On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3$ . In degrees, what was the low temperature in Lincoln that day?
$\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \te... | Let $L$ be the low temperature. The high temperature is $L+16$ . The average is $\frac{L+(L+16)}{2}=3$ . Solving for $L$ , we get $L=\boxed{5}$ | 5 |
4,578 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_2 | 1 | Mr. Green measures his rectangular garden by walking two of the sides and finds that it is $15$ steps by $20$ steps. Each of Mr. Green's steps is $2$ feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?
$\textbf{(A)}\... | Since each step is $2$ feet, his garden is $30$ by $40$ feet. Thus, the area of $30(40) = 1200$ square feet. Since he is expecting $\frac{1}{2}$ of a pound per square foot, the total amount of potatoes expected is $1200 \times \frac{1}{2} = \boxed{600}$ | 600 |
4,579 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_3 | 1 | When counting from $3$ to $201$ $53$ is the $51^{st}$ number counted. When counting backwards from $201$ to $3$ $53$ is the $n^{th}$ number counted. What is $n$
$\textbf{(A)}\ 146 \qquad \textbf{(B)}\ 147 \qquad \textbf{(C)}\ 148 \qquad \textbf{(D)}\ 149 \qquad \textbf{(E)}\ 150$ | Note that $n$ is equal to the number of integers between $53$ and $201$ , inclusive. Thus, $n=201-53+1=\boxed{149}$ | 149 |
4,580 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | 1 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | Let Ray and Tom drive 40 miles. Ray's car would require $\frac{40}{40}=1$ gallon of gas and Tom's car would require $\frac{40}{10}=4$ gallons of gas. They would have driven a total of $40+40=80$ miles, on $1+4=5$ gallons of gas, for a combined rate of $\frac{80}{5}=$ $\boxed{16}$ | 16 |
4,581 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | 2 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | Taking the harmonic mean of the two rates, we get \[\left(\frac{40^{-1} + 10^{-1}}{2}\right)^{-1} = \frac{2}{\frac{1}{40}+\frac{1}{10}} = \frac{2}{\frac{5}{40}} = \frac{2}{\frac{1}{8}} = \boxed{16}.\] | 16 |
4,582 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_4 | 3 | Ray's car averages $40$ miles per gallon of gasoline, and Tom's car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? $\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qqu... | Let the number of miles that Ray and Tom each drive be denoted by $m$ . Thus the number of gallons Ray's car uses can be represented by $\frac{m}{40}$ and the number of gallons that Tom's car uses can likewise be expressed as $\frac{m}{10}$ . Thus the total amount of gallons used by both cars can be expressed as $\frac... | 16 |
4,583 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_6 | 2 | Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$ . What is $x+y$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$ . Equality holds when the v... | 2 |
4,584 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | 1 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-... | 8 |
4,585 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | 2 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n... | 8 |
4,586 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_7 | 3 | Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $``1"$ , so Blair follows by saying $``1, 2"$ . Jo then says $``1, 2, 3"$ , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad ... | Let $T(n)$ denote the $n$ th triangle number. Then, observe that the $T(n)$ th number said is $n$ . It follows that the $55$ th number is $10$ (as $55 = T(10)$ ). Thus, the $53$ rd number is $10 - 2 = 8$ , which is answer choice $\boxed{8}$ | 8 |
4,587 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_9 | 1 | What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | Looking at the prime numbers under $12$ , we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of $2$ $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$ , and $\lef... | 8 |
4,588 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_10 | 1 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges... | If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first ... | 103 |
4,589 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_10 | 2 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges... | We can approach this problem by assuming he goes to the red booth first. You start with $75 \text{R}$ and $75 \text{B}$ and at the end of the first booth, you will have $1 \text{R}$ and $112 \text{B}$ and $37 \text{S}$ . We now move to the blue booth, and working through each booth until we have none left, we will end ... | 103 |
4,590 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_10 | 3 | Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges... | Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \[R(x,y)=-2x+y+75\] \[B(x,y)=x-3y+75\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins... | 103 |
4,591 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_12 | 1 | Cities $A$ $B$ $C$ $D$ , and $E$ are connected by roads $\widetilde{AB}$ $\widetilde{AD}$ $\widetilde{AE}$ $\widetilde{BC}$ $\widetilde{BD}$ $\widetilde{CD}$ , and $\widetilde{DE}$ . How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities mor... | Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$ , there is only one possible path to take out of cities $C$ or $E$ .
So the diagram is as follows:
[asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.0... | 16 |
4,592 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_13 | 1 | The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$ . Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the tw... | Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ $a+d$ $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$
For the sake of simplicity, lets rename the angles of each similar t... | 240 |
4,593 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_14 | 1 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 8... | Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$ . Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$ . Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$ . To minimize, let one... | 104 |
4,594 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_14 | 2 | Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$ . What is the smallest possible value of $N$
$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 8... | WLOG, let $a_i$ $b_i$ be the sequences with $a_1<b_1$ . Then \[N=5a_1+8a_2=5b_1+8b_2\] or \[5a_1+8a_2=5(a_1+c)+8(a_2-d)\] for some natural numbers $c$ $d$ . Thus $5c=8d$ . To minimize $c$ and $d$ , we have $(c,d)=(8,5)$ , or \[5a_1+8a_2=5(a_1+8)+8(a_2-5).\] To minimize $a_1$ and $b_1$ , we have $(a_1,b_1)=(0,0+c)=(0,8)... | 104 |
4,595 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_15 | 1 | The number $2013$ is expressed in the form
where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | The prime factorization of $2013$ is $61\cdot11\cdot3$ . To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$ . Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will le... | 2 |
4,596 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_16 | 1 | Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$ . The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$
$\textbf{(A)}\ 0 ... | The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$ , and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure... | 0 |
4,597 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_16 | 2 | Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$ . The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$
$\textbf{(A)}\ 0 ... | The extreme case, that results in the minimum and/or maximum, would probably be a pentagon that approaches a degenerate pentagon. However, due to the way the problem is phrased, we know there exists a minimum and maximum; therefore, we can reasonably assume that the star's perimeter is constant, and answer with $\boxed... | 0 |
4,598 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | 1 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | Since $\angle{AFB}=\angle{ADB}=90^{\circ}$ , quadrilateral $ABDF$ is cyclic. It follows that $\angle{ADE}=\angle{ABF}$ , so $\triangle ABF \sim \triangle ADE$ are similar. In addition, $\triangle ADE \sim \triangle ACD$ . We can easily find $AD=12$ $BD = 5$ , and $DC=9$ using Pythagorean triples.
So, the ratio of the l... | 21 |
4,599 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | 2 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | From solution 1, we know that $AD = 12$ and $DC = 9$ . Since $\triangle ADC \sim \triangle DEC$ , we can figure out that $DE = \frac{36}{5}$ . We also know what $AC$ is so we can figure what $AE$ is: $AE = 15 - \frac{27}{5} = \frac{48}{5}$ . Quadrilateral $ABDF$ is cyclic, implying that $\angle{B} + \angle{DFA}$ = 180°... | 21 |
4,600 | https://artofproblemsolving.com/wiki/index.php/2013_AMC_12B_Problems/Problem_19 | 3 | In triangle $ABC$ $AB=13$ $BC=14$ , and $CA=15$ . Distinct points $D$ $E$ , and $F$ lie on segments $\overline{BC}$ $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD}\perp\overline{BC}$ $\overline{DE}\perp\overline{AC}$ , and $\overline{AF}\perp\overline{BF}$ . The length of segment $\overli... | If we draw a diagram as given, but then add point $G$ on $\overline{BC}$ such that $\overline{FG}\perp\overline{BC}$ in order to use the Pythagorean theorem, we end up with similar triangles $\triangle{DFG}$ and $\triangle{DCE}$ . Thus, $FG=\tfrac35x$ and $DG=\tfrac45x$ , where $x$ is the length of $\overline{DF}$ . Us... | 21 |
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