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 |
|---|---|---|---|---|---|
6,901 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | 1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | Note that the four numbers to multiply are symmetric with the center at $29.5$ .
Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$ $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$ | 869 |
6,902 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | 4 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | Similar to Solution 1 above, call the consecutive integers $\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)$ to make use of symmetry. Note that $n$ itself is not an integer - in this case, $n = 29.5$ . The expression becomes $\sqrt{\left(n-\frac{3}{2}\rig... | 869 |
6,903 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | 5 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | We have $(31)(30)(29)(28)+1=755161.$ Since the alternating sum of the digits $7-5+5-1+6-1=11$ is divisible by $11,$ we conclude that $755161$ is divisible by $11.$
We evaluate the original expression by prime factorization: \begin{align*} \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ &=\sqrt{11\cdot68651} \\ &=\sqrt{11^2... | 869 |
6,904 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | 6 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | The last digit under the radical is $1$ , so the square root must either end in $1$ or $9$ , since $x^2 = 1\pmod {10}$ means $x = \pm 1$ . Additionally, the number must be near $29 \cdot 30 = 870$ , narrowing the reasonable choices to $869$ and $871$
Continuing the logic, the next-to-last digit under the radical is t... | 869 |
6,905 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_2 | 1 | Ten points are marked on a circle . How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices | Any subset of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are $2^{10} = 1024$ total subsets of a ten-member set , but of these ${10 \choose 0} = 1$ have 0 members, ${10 \choose 1} = 10$ have 1 member and ${10 \choose 2} ... | 968 |
6,906 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | 1 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | Repeating decimals represent rational numbers . To figure out which rational number, we sum an infinite geometric series $0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$ . Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$ . Since 75... | 750 |
6,907 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | 2 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | To get rid of repeating decimals, we multiply the equation by 1000. We get $\frac{1000n}{810} = d25.d25d25...$ We subtract the original equation from the second to get $\frac{999n}{810}=d25$ We simplify to $\frac{37n}{30} = d25$ Since $\frac{37n}{30}$ is an integer, $n=(30)(5)(2k+1)$ because $37$ is relatively prime to... | 750 |
6,908 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | 3 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | Similar to Solution 2, we start off by writing that $\frac{1000n}{810} = d25.d25d25 \dots$ .Then we subtract this from the original equation to get:
\[\frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30\]
Since n is an integer, we have that $37 \mid d25 \cdot 30$
Since $37$ is ... | 750 |
6,909 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | 4 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | Write out these equations:
$\frac{n}{810} = \frac{d25}{999}$
$\frac{n}{30} = \frac{d25}{37}$
$37n = 30(d25)$
Thus $n$ divides 25 and 30. The only solution for this under 1000 is $\boxed{750}$ | 750 |
6,910 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4 | 1 | If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$ | Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$ . Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ de... | 675 |
6,911 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4 | 2 | If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$ | Let $b$ $c$ $d$ , and $e$ equal $a+1$ $a+2$ $a+3$ , and $a+4$ , respectively. Call the square and cube $k^2$ and $m^3$ , where both k and m are integers. Then:
$5a + 10 = m^3$
Now we know $m^3$ is a multiple of 125 and $m$ is a multiple of 5. The lower $m$ is, the lower the value of $c$ will be. Start from 5 and add 5 ... | 675 |
6,912 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4 | 3 | If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$ | Let the numbers be $a,a+1,a+2,a+3,a+4.$ When then know $3a+6$ is a perfect cube and $5a+10$ is perfect cube. Since $5a+10$ is divisible by $5$ we know that $5a+10 = (5k)^3$ since otherwise we get a contradiction. This means $a = 25k^3 - 2$ in which plugging into the other expression we know $3(25k^3 - 2) + 6 = 75k^3$ i... | 675 |
6,913 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_4 | 4 | If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube , what is the smallest possible value of $c$ | (This is literally a combination of 1 and 3)
Since $a$ $b$ $c$ $d$ , and $e$ are consecutive, $a = c-2$ $b = c-1$ $c=c$ $d = c+1$ , and $e = c+2$
Because $b+c+d = 3c$ is a perfect square, and $a+b+c+d+e = 5c$ is a perfect cube, we can express $c$ as $c = 3^{n} \cdot 5^{k}$
Now, by the problem's given information,
$k \e... | 675 |
6,914 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_5 | 1 | When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$ , in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$ | Denote the probability of getting a heads in one flip of the biased coin as $h$ . Based upon the problem, note that ${5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3$ . After canceling out terms, we get $1 - h = 2h$ , so $h = \frac{1}{3}$ . The answer we are looking for is ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3... | 283 |
6,915 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_5 | 2 | When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$ , in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$ | Denote the probability of getting a heads in one flip of the biased coins as $h$ and the probability of getting a tails as $t$ . Based upon the problem, note that ${5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3$ . After cancelling out terms, we end up with $t = 2h$ . To find the probability getting $3$ heads, we need to... | 283 |
6,916 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6 | 1 | Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leave... | Label the point of intersection as $C$ . Since $d = rt$ $AC = 8t$ and $BC = 7t$ . According to the law of cosines
\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100... | 160 |
6,917 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6 | 2 | Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leave... | Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\overline{AB}$ . Letting this line be the $x$ -axis, we can reflect $B$ over the $x$ -axis to get $B'$ . As reflections preserve length, $B'X = XB$
We then draw lines $BB'$ and $PB'$ ... | 160 |
6,918 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_6 | 3 | Two skaters, Allie and Billie, are at points $A$ and $B$ , respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$ . At the same time Allie leaves $A$ , Billie leave... | We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$ . Then we can produce the following diagram:
[asy] draw((0,0)--(100,0)--(80,139)--cycle); label("8x",(0,0)--(80,139),NW); label("7x",(100,0)--(80,139),NE); ... | 160 |
6,919 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_7 | 1 | If the integer $k$ is added to each of the numbers $36$ $300$ , and $596$ , one obtains the squares of three consecutive terms of an arithmetic series. Find $k$ | Call the terms of the arithmetic progression $a,\ a + d,\ a + 2d$ , making their squares $a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2$
We know that $a^2 = 36 + k$ and $(a + d)^2 = 300 + k$ , and subtracting these two we get $264 = 2ad + d^2$ (1). Similarly, using $(a + d)^2 = 300 + k$ and $(a + 2d)^2 = 596 + k$ , subtract... | 925 |
6,920 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_7 | 2 | If the integer $k$ is added to each of the numbers $36$ $300$ , and $596$ , one obtains the squares of three consecutive terms of an arithmetic series. Find $k$ | Since terms in an arithmetic progression have constant differences, \[\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}\] \[\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}\] \[\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}\] \[\implies 2k+568=2\sqrt{(596+k)(36+k)}\] \[\implies k+284=\sqrt{(596+k)(36+k)}\] \[\impl... | 925 |
6,921 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 1 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | Note that each given equation is of the form \[f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7\] for some $k\in\{1,2,3\}.$
When we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ \[f(k)=ak^2+bk+c,\] where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x... | 334 |
6,922 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 2 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | For simplicity purposes, we number the given equations $(1),(2),$ and $(3),$ in that order. Let \[16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \hspace{29.5mm}(4)\] Subtracting $(1)$ from $(2),$ subtracting $(2)$ from $(3),$ and subtracting $(3)$ from $(4),$ we obtain the following equations, respectively: \begin{align... | 334 |
6,923 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 3 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:
[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=1; i<=10; ++i) { label("\boldmath{$"+string(i^2)+"$}",(i-1,0)); } for (real i=1; i<=9; ++i) { label("$"+s... | 334 |
6,924 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 4 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | Notice that we may rewrite the equations in the more compact form as: \begin{align*} \sum_{i=1}^{7}i^2x_i&=c_1, \\ \sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\ \sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\ \sum_{i=1}^{7}(i+3)^2x_i&=c_4, \end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find.
Now consider the pol... | 334 |
6,925 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 5 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | The idea is to multiply the first, second and third equations by $a,b,$ and $c,$ respectively.
We can only consider the coefficients of $x_1,x_2,$ and $x_3:$ \begin{align} a+4b+9c&=16, \\ 4a+9b+16c&=25, \\ 9a+16b+25c&=36. \end{align} Subtracting $(1)$ from $(2),$ we get \[3a+5b+7c=9. \hspace{15mm}(4)\] Subtracting $3\c... | 334 |
6,926 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 6 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$ . Thus, we have \begin{align*} x_1+4x_2+9x_3&=1,\\ 4x_1+9x_2+16x_3&=12,\\ 9x_1+16x_2+25x_3&=123.\\ \end{align*} Grinding this out, we have $(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)$ which gives $\boxed{334}$ as our final answer. | 334 |
6,927 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_8 | 7 | Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | Let $s_n = n^2$ be the sequence of perfect squares.
By either expanding or via finite differences, one can prove the miraculous recursion \[s_n = 3s_{n-1} - 3s_{n-2} + s_{n-3}.\] Hence, the answer is simply \[3 \cdot 123 - 3 \cdot 12 + 1 = \boxed{334}.\] | 334 |
6,928 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | 1 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | Taking the given equation modulo $2,3,$ and $5,$ respectively, we have \begin{align*} n^5&\equiv0\pmod{2}, \\ n^5&\equiv0\pmod{3}, \\ n^5&\equiv4\pmod{5}. \end{align*} By either Fermat's Little Theorem (FLT) or inspection, we get \begin{align*} n&\equiv0\pmod{2}, \\ n&\equiv0\pmod{3}, \\ n&\equiv4\pmod{5}. \end{align*}... | 144 |
6,929 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | 2 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | Note that $n$ is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know $n^5\equiv n\pmod{5}.$ Hence, \[n\equiv3+0+4+2\equiv4\pmod{5}.\] Continuing, we examine the equation modulo $3,$ \[n\equiv1-1+0+0\equiv0\pmod{3}.\] Thus, $n$ is divisible by three and leaves a remainder of... | 144 |
6,930 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | 3 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, $n^5\equiv n\pmod{5},$ and it is easy to see that $n^5\equiv n\pmod 2.$ Therefore, $133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10},$ so the last digit of $n$ is $4.$
We notice that $133,110,84,$ and $27$ are all ver... | 144 |
6,931 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | 4 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | In this solution we take advantage of the large numbers and utilize parity properties to give us a very good guess at the answer. The units digits of $133^5, 110^5, 84^5, 27^5$ are $3, 0, 4, 7,$ respectively, so the units digit of $n^5$ is $4.$ This tells us $n$ is even. Since we are dealing with enormous numbers, $n$ ... | 144 |
6,932 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | 5 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | First, we take mod $2$ on both sides to get $n^5\equiv 0\pmod{2}\implies n\equiv 0\pmod{2}$ . Mod $3$ gives $n^5\equiv 0\pmod{3}\implies n\equiv 0\pmod{3}$ . Also, mod $5$ gives $n^5\equiv -1\pmod{5}\implies n\equiv -1\pmod{5}$ (by FLT). Finally, note that mod $7$ gives $n^5\equiv 2\pmod{7}\implies n^{-1}\equiv 2\pmod{... | 144 |
6,933 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_9 | 6 | One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$ | We have \[n^5 = 133^5 + 110^5 + 84^5 +27^5 = 61917364224,\] for which $n = \sqrt [5]{61917364224} = \boxed{144}.$ | 144 |
6,934 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 1 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | We draw the altitude $h$ to $c$ , to get two right triangles
Then $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$ , from the definition of the cotangent
Let $K$ be the area of $\triangle ABC.$ Then $h=\frac{2K}{c}$ , so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}$
By identical logic, we can find similar expressions for the sums of... | 994 |
6,935 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 2 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | We start as in solution 1, though we'll write $A$ instead of $K$ for the area. Now we evaluate the numerator:
\[\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}\]
From the Law of Cosines and the sine area formula,
\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\ \sin{\gamma}&= \frac{2A}{ab}\\ \cot{\gamma}&= \frac{\cos ... | 994 |
6,936 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 3 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | \begin{align*} \cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac {\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}\\ &= \frac {\sin{(\alpha + \beta)}}{\sin{\alpha}\sin{\beta}} = \frac {\sin{\gamma}}{\sin{\alpha}\sin{\beta}} \end{align*... | 994 |
6,937 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 4 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | Use Law of cosines to give us $c^2=a^2+b^2-2ab\cos(\gamma)$ or therefore $\cos(\gamma)=\frac{994c^2}{ab}$ . Next, we are going to put all the sin's in term of $\sin(a)$ . We get $\sin(\gamma)=\frac{c\sin(a)}{a}$ . Therefore, we get $\cot(\gamma)=\frac{994c}{b\sin a}$
Next, use Law of Cosines to give us $b^2=a^2+c^2-... | 994 |
6,938 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 5 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | Let $\gamma$ be $(180-\alpha-\beta)$
$\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}$
WLOG, assume that $a$ and $c$ are legs... | 994 |
6,939 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 6 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | As in Solution 1, drop an altitude $h$ to $c$ . Let $h$ meet $c$ at $P$ , and let $AP = x, BP = y$
Then, $\cot{\alpha} = \frac{1}{\tan{\alpha}} = \frac{x}{h}$ $\cot{\beta} = \frac{1}{\tan{\beta}} = \frac{y}{h}$ . We can calculate $\cot{\gamma}$ using the tangent addition formula , after noticing that $\cot{\gamma} = \f... | 994 |
6,940 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_10 | 7 | Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find | Since no additional information is given, we can assume that triangle ABC is right with the right angle at B.
We can use the Pythagorean theorem to say \[c^2+a^2=b^2\] We can now solve for $a$ in terms of $c$
\[c^2+a^2=1989c^2-a^2\] \[a^2=994c^2\] \[a=\sqrt{994}c\]
Using the definition of cotangent
\[cot(A)=\frac{c}{a}... | 994 |
6,941 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_11 | 1 | A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$ ? (For real $x$ $\lfloor x\rfloor... | Let the mode be $x$ , which we let appear $n > 1$ times. We let the arithmetic mean be $M$ , and the sum of the numbers $\neq x$ be $S$ . Then \begin{align*} D &= \left|M-x\right| = \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right| \end{align*} As $S$ is essentially independen... | 947 |
6,942 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_11 | 2 | A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$ ? (For real $x$ $\lfloor x\rfloor... | With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\underbrace{1,1,1...1,}_\text{n times}\underbrace{2,2,2...2,}_\text{n times}\underbrace{3,3,3...3,}_\text{n times}........\underbrace{1000,1000,1000....}_\text{n+1 times}$ And, we need to fi... | 947 |
6,943 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_13 | 1 | Let $S$ be a subset of $\{1,2,3,\ldots,1989\}$ such that no two members of $S$ differ by $4$ or $7$ . What is the largest number of elements $S$ can have? | We first show that we can choose at most 5 numbers from $\{1, 2, \ldots , 11\}$ such that no two numbers have a difference of $4$ or $7$ . We take the smallest number to be $1$ , which rules out $5,8$ . Now we can take at most one from each of the pairs: $[2,9]$ $[3,7]$ $[4,11]$ $[6,10]$ . Now, $1989 = 180\cdot 11 + 9$... | 905 |
6,944 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 1 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | Let $[RST]$ be the area of polygon $RST$ . We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$ , and line $XP$ intersects line $YZ$ at point $L$ , then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$
This is true because triangles $XPY$ and $YPL$ have their areas in ratio $XP:PL$ ... | 108 |
6,945 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 2 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points . We immediately see that $w_E = 3$ $w_B = 1$ , and $w_A = w_D = 2$ . Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C = 1$ and $w_F = 3$ . Thus, $CP = 15$ and $PF =... | 108 |
6,946 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 3 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | Using a different form of Ceva's Theorem , we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$
Solving $4y = x + y$ and $x + y = 20$ , we obtain $x = CP = 15$ and $y = FP = 5$
Let $Q$ be the point on $AB$ such that $FC \parallel QD$ .
Since $AP = PD$ an... | 108 |
6,947 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 4 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | First, let $[AEP]=a, [AFP]=b,$ and $[ECP]=c.$ Thus, we can easily find that $\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.$ Now, $\frac{[ABP]}{[BPD]}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.$ In the same manner, we find that $[CPD]=a+c.$ Now, we can find that $\frac{[BPC]}{[PEC]}=\frac{9}{3... | 108 |
6,948 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 5 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | Firstly, since they all meet at one single point, denoting the mass of them separately. Assuming $M(A)=6;M(D)=6;M(B)=3;M(E)=9$ ; we can get that $M(P)=12;M(F)=9;M(C)=3$ ; which leads to the ratio between segments, \[\frac{CE}{AE}=2;\frac{BF}{AF}=2;\frac{BD}{CD}=1.\] Denoting that $CE=2x;AE=x; AF=y; BF=2y; CD=z; DB=z.$
... | 108 |
6,949 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 6 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | In Solution 5, instead of finding all of $x, y, z$ , we only need $y, z$ . This is because after we solve for $y, z$ , we can notice that $\triangle BAD$ is isosceles with $AB = BD$ . Because $P$ is the midpoint of the base, $BP$ is an altitude of $\triangle BAD$ . Therefore, $[BAD] = \frac{(AD)(BP)}{2} = \frac{12 \cdo... | 108 |
6,950 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 7 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | Set $AF=x,$ and use mass points to find that $PF=5$ and $BF=2x.$ Using Stewart's Theorem on $APB,$ we find that $AB=3\sqrt{13}.$ Then we notice that $APB$ is right, which means the area of $APB$ is $27.$ Because $CF=4\cdot PF,$ the area of $ABC$ is $4$ times the area of $APB,$ which means the area of $ABC=4\cdot 27=\bo... | 108 |
6,951 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_15 | 9 | Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$ | We start with mass points as in Solution 2, and receive $BF:AF = 2$ $BD:CD = 1$ $CE:AE = 2$ Law of Cosines on triangles $ADB$ and $ADC$ with $\theta = \angle ADB$ and $BD=DC=x$ gives \[36+x^2-12x\cos \theta = 81\] \[36+x^2-12x\cos (180-\theta) = 36+x^2+12x\cos \theta = 225\] Adding them: $72+2x^2=306 \implies x=3\sqrt{... | 108 |
6,952 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_1 | 1 | One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination . Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many ... | Currently there are ${10 \choose 5}$ possible combinations.
With any integer $x$ from $1$ to $9$ , the number of ways to choose a set of $x$ buttons is $\sum^{9}_{k=1}{10 \choose k}$ .
Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$ .
So the number of additional combinations is just $2^{10}-{10\choose 0... | 770 |
6,953 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_2 | 1 | For any positive integer $k$ , let $f_1(k)$ denote the square of the sum of the digits of $k$ . For $n \ge 2$ , let $f_n(k) = f_1(f_{n - 1}(k))$ . Find $f_{1988}(11)$ | We see that $f_{1}(11)=4$
$f_2(11) = f_1(4)=16$
$f_3(11) = f_1(16)=49$
$f_4(11) = f_1(49)=169$
$f_5(11) = f_1(169)=256$
$f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$ | 169 |
6,954 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_3 | 2 | Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$ | We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2.
\begin{align*} {\log_2 (\frac{1}{3}\log_2 x)} &= \frac{1}{3}{\log_2 (\log_2 x)}\\ {\log_2 x = y}\\ {\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ {3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\ {\log_2 (\f... | 27 |
6,955 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_3 | 3 | Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$ | First we have \begin{align*} \log_2(\log_8x)&=\log_8(\log_2x)\\ \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 \end{align*} Changing the base in the numerator yields \begin{align*} \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ \end{align*} Using the property $\frac{\lo... | 27 |
6,956 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_3 | 4 | Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$ | Say that $\log_{2^3}x=a$ and $\log_2x=b$ so we have $\log_2a=\log_{2^3}b$ . And we want $b^2$
$\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.$
Because $3a=b$ (as $2^{3a}=x$ and $2^b=x$ from our setup), we have that
$b^{1/3}=\frac{b}{3}\\ b^{-2/3}=\frac13... | 27 |
6,957 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_5 | 1 | Let $m/n$ , in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$ . Find $m + n$ | $10^{99} = 2^{99}5^{99}$ , so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$ ; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$ . Our probability is $\frac{m}{n}... | 634 |
6,958 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6 | 1 | It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png | Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc.
Label the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for $a$ an... | 142 |
6,959 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6 | 2 | It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png | First, let $a =$ the number to be placed in the first column, fourth row. Let $b =$ the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of $a$ and $b$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\ \hline 3a & & & & \\ \hline 2a & & & & \\ \hl... | 142 |
6,960 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_6 | 3 | It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png | We begin with the table that was given to us and add in the following arithmetic progression on the bottom:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline & & & * & \\ \hline & 74 & & & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$
Since all the rows and columns satisfy an ... | 142 |
6,961 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_7 | 1 | In triangle $ABC$ $\tan \angle CAB = 22/7$ , and the altitude from $A$ divides $BC$ into segments of length 3 and 17. What is the area of triangle $ABC$ | Call $\angle BAD$ $\alpha$ and $\angle CAD$ $\beta$ . So, $\tan \alpha = \frac {17}{h}$ and $\tan \beta = \frac {3}{h}$ . Using the tangent addition formula $\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$ , we get $\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 -... | 110 |
6,962 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8 | 1 | The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$ | Let $z = x+y$ . By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\ f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x). \end{align*} Using the properties of $f,$ we have \begin{align*} f(14,52) &= \frac{52}{38} \cdot f(14,38) \\ &= ... | 364 |
6,963 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8 | 2 | The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$ | Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one.
N... | 364 |
6,964 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_8 | 3 | The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$ | Notice that $f(x,y) = \mathrm{lcm}(x,y)$ satisfies all three properties:
For the first two properties, it is clear that $\mathrm{lcm}(x,x) = x$ and $\mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)$
For the third property, using the identities $\gcd(x,y) \cdot \mathrm{lcm}(x,y) = x\cdot y$ and $\gcd(x,x+y) = \gcd(x,y)$ gives \beg... | 364 |
6,965 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9 | 1 | Find the smallest positive integer whose cube ends in $888$ | $n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8$ and $n^3 \equiv 13 \pmod{125}$ $n \equiv 2 \pmod 5$ due to the last digit of $n^3$ . Let $n = 5a + 2$ . By expanding, $125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}$
By looking at the last digit again, we see $a \equiv 3 ... | 192 |
6,966 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9 | 2 | Find the smallest positive integer whose cube ends in $888$ | Let $x^3 = 1000a + 888$ . We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$ , where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$ . Taking mod $5$ $25$ , and $125$ , we get $y^3 \equiv 1\pmod 5$ $y^3 \equiv 11\pmod{25}$ , and $y^3 \equiv 111\pmod{125}$
We ... | 192 |
6,967 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9 | 3 | Find the smallest positive integer whose cube ends in $888$ | Let this integer be $x.$ Note that \[x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.\] We wish to find the residue of $x$ mod $125.$ Note that \[x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}\] using our congruence in mod $5.$ The residue that works must also satisfy $x^3 \equiv 13... | 192 |
6,968 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_9 | 4 | Find the smallest positive integer whose cube ends in $888$ | This number is in the form of $10k+2$ , after binomial expansion, we only want $600k^2+120k\equiv 880 \pmod{1000}$ . We realize that $600,120$ are both multiples of $8$ , we only need that $600k^2+120k \equiv 5\pmod{125}$ , so we write $600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}$
T... | 192 |
6,969 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10 | 1 | convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face | The polyhedron described looks like this, a truncated cuboctahedron.
The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$ . We must now subtract out those segments that lie along an edge or a face.
Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octa... | 840 |
6,970 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10 | 2 | convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face | We first find the number of vertices on the polyhedron:
There are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. $\text{vertices} = \frac{12 \cdot 4 + 8 \cdot 6 + 6 \cdot 8}{3}=48$
We know that all vertices look t... | 840 |
6,971 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10 | 3 | convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face | Since at each vertex one square, one hexagon, and one octagon meet, then there are a total of $12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$ vertices. This means that for each segment we have $48$ choices of vertices for the first endpoint of the segment.
Since each vertex is the meeting point of a square, octagon, and hexa... | 840 |
6,972 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_10 | 4 | convex polyhedron has for its faces 12 squares , 8 regular hexagons , and 6 regular octagons . At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face | In the same ways as above, we find that there are 48 vertices. Now, notice that there are $\binom{48}{2}$ total possible ways to choose two vertices. However, we must remove the cases where the segments do not lie in the interior of the polyhedron. We get
\[\binom{48}{2}-12\binom{4}{2}-8\binom{6}{2}-6\binom{8}{2}=768\]... | 840 |
6,973 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_11 | 1 | Let $w_1, w_2, \dots, w_n$ be complex numbers . A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$ $w_2 = - 7 + 64i$ $w_3 = - 9 + 200i... | We know that
$\sum_{k=1}^5 w_k = 3 + 504i$
And because the sum of the 5 $z$ 's must cancel this out,
$\sum_{k=1}^5 z_k = 3 + 504i$
We write the numbers in the form $a + bi$ and we know that
$\sum_{k=1}^5 a_k = 3$ and $\sum_{k=1}^5 b_k = 504$
The line is of equation $y=mx+3$ . Substituting in the polar coordinates, we h... | 163 |
6,974 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_11 | 2 | Let $w_1, w_2, \dots, w_n$ be complex numbers . A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$ $w_2 = - 7 + 64i$ $w_3 = - 9 + 200i... | The mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\frac{3}{5}, \frac{504i}{5})$ . Since we now have two points, namely that one and $(0, 3i)$ , we can simply find the slope between them, which is $\boxed{163}$ by the g... | 163 |
6,975 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12 | 1 | Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$
1988 AIME-12.png | Call the cevians AD, BE, and CF. Using area ratios ( $\triangle PBC$ and $\triangle ABC$ have the same base), we have:
$\frac {d}{a + d} = \frac {[PBC]}{[ABC]}$
Similarily, $\frac {d}{b + d} = \frac {[PCA]}{[ABC]}$ and $\frac {d}{c + d} = \frac {[PAB]}{[ABC]}$
Then, $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + ... | 441 |
6,976 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12 | 2 | Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$
1988 AIME-12.png | Let $A,B,C$ be the weights of the respective vertices. We see that the weights of the feet of the cevians are $A+B,B+C,C+A$ . By mass points , we have that: \[\dfrac{a}{3}=\dfrac{B+C}{A}\] \[\dfrac{b}{3}=\dfrac{C+A}{B}\] \[\dfrac{c}{3}=\dfrac{A+B}{C}\]
If we add the equations together, we get $\frac{a+b+c}{3}=\frac{A^2... | 441 |
6,977 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12 | 3 | Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$
1988 AIME-12.png | You can use mass points to derive $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d}=1.$ Plugging it in yields $\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1.$ We proceed as we did in Solution 1 - however, to make the equation look less messy, we do the substitution $a'=a+3,b'=b+3,c'=c+3.$
Then we have $\f... | 441 |
6,978 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_12 | 4 | Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$ $b$ $c$ , and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$
1988 AIME-12.png | A cool identity derived from Ceva's Theorem is that:
\[\frac{AP}{PA'}\frac{BP}{PB'}\frac{CP}{PC'} = 2 + \frac{AP}{PA'} + \frac{BP}{PB'} + \frac{CP}{PC'}\]
To see this, we use another Ceva's Theorem identity (sometimes attributed to Gergonne): $\frac{AP}{PA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}$ , and similarly for cevians ... | 441 |
6,979 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13 | 1 | Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ | Let $F_n$ represent the $n$ th number in the Fibonacci sequence. Therefore, \begin{align*} x^2 - x - 1 = 0&\Longrightarrow x^n = F_n(x), \ n\in N \\ &\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N. \end{align*} The above uses the similarity between the Fibonacci recursion|recursive definition, $F_{n+2} - F... | 987 |
6,980 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13 | 2 | Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ | We can long divide and search for a pattern; then the remainder would be set to zero to solve for $a$ . Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is \[(F_{16}b + F_{17}a)x + F_{15}b + F_{16}... | 987 |
6,981 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13 | 3 | Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ | Trying to divide $ax^{17} + bx^{16} + 1$ by $x^2-x-1$ would be very tough, so let's try to divide using smaller degrees of $x$ . Doing $\frac{ax^3+bx^2+1}{x^2-x-1}$ , we get the following systems of equations: \begin{align*} a+b &= -1, \\ 2a+b &= 0. \end{align*} Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$ \begin{alig... | 987 |
6,982 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13 | 4 | Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ | Let's work backwards! Let $F(x) = ax^{17} + bx^{16} + 1$ and let $P(x)$ be the polynomial such that $P(x)(x^2 - x - 1) = F(x)$
Clearly, the constant term of $P(x)$ must be $- 1$ . Now, we have \[(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1),\] where $c_{i}$ is some coefficient. However, since $F(x)$ has n... | 987 |
6,983 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13 | 6 | Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ | We are given that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1,$ so the roots of $x^2 - x - 1$ must also be roots of $ax^{17} + bx^{16} + 1.$
Let $x=r$ be a root of $x^2 - x - 1$ so that $r^2 - r - 1 = 0,$ or $r^2 = r + 1.$ It follows that \[ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)\] Note that \begin{... | 987 |
6,984 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_13 | 7 | Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ | For simplicity, let $f(x) =ax^{17} + bx^{16} + 1$ and $g(x) = x^2-x-1$ . Notice that the roots of $g(x)$ are also roots of $f(x)$ . Let these roots be $u,v$ . We get the system \begin{align*} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{align*} If we multiply the first equation by $v^{16}$ and the se... | 987 |
6,985 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_15 | 1 | In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss deliver... | Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$ , or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given that $8$ had b... | 704 |
6,986 | https://artofproblemsolving.com/wiki/index.php/1988_AIME_Problems/Problem_15 | 2 | In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss deliver... | $1 \cdot 2 + 7 \cdot 3 + 21 \cdot 4 + 35 \cdot 5 + 35 \cdot 6 + 21 \cdot 7 + 7 \cdot 8 + 1 \cdot 9$ $=1 \cdot (2+9) + 7 \cdot (3+8) + 21 \cdot (4+7) + 35 \cdot (5+6) = 64 \cdot 11 = \boxed{704}$ | 704 |
6,987 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_1 | 1 | An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$ | Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}$ | 300 |
6,988 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_2 | 1 | What is the largest possible distance between two points , one on the sphere of radius 19 with center $(-2,-10,5)$ and the other on the sphere of radius 87 with center $(12,8,-16)$ | The distance between the two centers of the spheres can be determined via the distance formula in three dimensions: $\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31$ . The largest possible distance would be the sum of the two radii and the distance between the two centers, making it ... | 137 |
6,989 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_3 | 1 | By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers? | Let $p(n)$ denote the product of the distinct proper divisors of $n$ . A number $n$ is nice in one of two instances:
We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form $n = pqr$ (with ... | 182 |
6,990 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_4 | 1 | Find the area of the region enclosed by the graph of $|x-60|+|y|=\left|\frac{x}{4}\right|.$ | 1987 AIME-4.png
Since $|y|$ is nonnegative $\left|\frac{x}{4}\right| \ge |x - 60|$ . Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60$ . Thus, $48 \le x \le 80$ . The maximum and minimum y value is when $|x - 60| = 0$ , which is when $x = 60$ and $y = \pm 15$ . Since ... | 480 |
6,991 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_5 | 1 | Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$ | If we move the $x^2$ term to the left side, it is factorable with Simon's Favorite Factoring Trick
\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]
$507$ is equal to $3 \cdot 13^2$ . Since $x$ and $y$ are integers, $3x^2 + 1$ cannot equal a multiple of three. $169$ doesn't work either, so $3x^2 + 1 = 13$ , and $x^2 = 4$ . This leav... | 588 |
6,992 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6 | 1 | Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.
AIME 1987 Problem 6.png | Since $XY = WZ$ $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$ . This number is also equal to one quarter the area of the entire rectangle, which... | 193 |
6,993 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6 | 2 | Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.
AIME 1987 Problem 6.png | Let $YB=a$ $CZ=b$ $AX=c$ , and $WD=d$ . First we drop a perpendicular from $Q$ to a point $R$ on $BC$ so $QR=h$ . Since $XY = WZ$ and $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, the heights of the trapezoids are both $\frac{19}{2}$ .From here, we have that $[BYQZC]=\frac{a+h}{2}*19/2+\frac... | 193 |
6,994 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_6 | 3 | Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$ , and $PQ$ is parallel to $AB$ . Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm.
AIME 1987 Problem 6.png | Since $XY = YB + BC + CZ = ZW = WD + DA + AX$ . Let $a = AX + DW = BY + CZ$ . Since $2AB - 2a = XY = WZ$ , then $XY = AB - a$ .Let $S$ be the midpoint of $DA$ , and $T$ be the midpoint of $CB$ . Since the area of $PQWZ$ and $PQYX$ are the same, then their heights are the same, and so $PQ$ is equidistant from $AB$ and $... | 193 |
6,995 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_8 | 1 | What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$ | Flip the fractions and subtract one from all sides to yield \[\frac{7}{8}>\frac{k}{n}>\frac{6}{7}.\] Multiply both sides by $56n$ to get \[49n>56k>48n.\] This is equivalent to find the largest value of $n$ such that there is only one multiple of 56 within the open interval between $48n$ and $49n$ . If $n=112,$ then $9... | 112 |
6,996 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_8 | 2 | What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$ | Notice that in order for $k$ to be unique, $\frac{n}{n + k+ 1} \le \frac{8}{15}$ and $\frac{n}{n+ k-1} \ge \frac{7}{13}$ must be true. Solving these inequalities for $k$ yields $\frac{7}{6}(k-1) \le n \le \frac{8}{7}(k+1)$
Thus, we want to find $k$ such that $\frac{7}{6}(k-1)\le \frac{8}{7}(k+1)$ . Solving this inequal... | 112 |
6,997 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | 1 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume tha... | Let the total number of steps be $x$ , the speed of the escalator be $e$ and the speed of Bob be $b$
In the time it took Bob to climb up the escalator he saw 75 steps and also climbed the entire escalator. Thus the contribution of the escalator must have been an additional $x - 75$ steps. Since Bob and the escalator ... | 120 |
6,998 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | 3 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume tha... | Let $e$ and $b$ be the speeds of the escalator and Bob, respectively.
When Al was on his way down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have \[150(3b-e)=75(b+e)\] Solving ge... | 120 |
6,999 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | 4 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume tha... | Please understand the machinery of an escalator before proceeding to read this solution.
Let the number of steps that disappear at the top of the escalator equal $x.$ Assume that Al takes $3$ steps per second and that Bob takes $1$ step per second. Since Al counts $150$ steps, it takes him \[\frac{150}{3}=50\] seconds ... | 120 |
7,000 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_10 | 5 | Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume tha... | WLOG, let Al's speed be $15$ steps per second, so Bob's speed is $5$ steps per second. Then, Al was on the escalator for $\frac{150}{15}\ = 10$ seconds and Bob was on the escalator for $\frac{75}{5}\ = 15$ seconds. Let $r$ be the rate of the escalator, in steps per second. Then, the total amount of steps is $150 - 10r ... | 120 |
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