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 |
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7,001 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_11 | 1 | Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers | Let us write down one such sum, with $m$ terms and first term $n + 1$
$3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$
Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$ . However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot 3... | 486 |
7,002 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_11 | 2 | Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers | Proceed as in Solution 1 until it is noted that $m$ is a divisor of $2\cdot 3^{11}$ . The divisors of $2\cdot 3^{11}$ are $3^{1} , 2\cdot 3^{1} , 3^{2} , 2\cdot 3^{2} , \ldots , 2\cdot 3^{10} , 3^{11}$ . Note that the factors of $m(2n + m + 1)$ are of opposite parity (if $m$ is odd, then $(2n + m + 1)$ is even and vice... | 486 |
7,003 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_13 | 1 | A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the thi... | If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$ , one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21}, r_{... | 931 |
7,004 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | 1 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that \begin{align*} x^4 + 324 &= x^4 + 4\cdot 3^4 \\ &= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\lef... | 373 |
7,005 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | 2 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$
We factor $N^4+18^2$ by completing the square, then applying the difference of squares: \begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+1... | 373 |
7,006 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | 3 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | We rewrite $N$ to the polar form \[N=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem , we have \[N^4=r^4\operatorname{cis}(4\theta)=18^2(-1),\] from which
By the Factor T... | 373 |
7,007 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14 | 4 | Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | We rewrite $N$ to the rectangular form \[N=a+bi\] for some real numbers $a$ and $b.$
Note that $N^2=\pm18i,$ so there are two cases:
By the Factor Theorem , we get \begin{align*} N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N... | 373 |
7,008 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | 1 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | 1987 AIME-15a.png
Because all the triangles in the figure are similar to triangle $ABC$ , it's a good idea to use area ratios . In the diagram above, $\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.$ Hence, $T_3 = \frac {440}{441}T_1$ and $T_4 = \frac {440}{441}T_2$ . Additionally, the area of triangle $ABC$ is... | 462 |
7,009 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | 2 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | Let $\tan\angle ABC = x$ . Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$ . Using the second square, $AB=\sqrt{440}(1+x+x^{-1})$ . We have $AC^2+CB^2=AB^2$ , or \[441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).\] Rearranging and letting $u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $... | 462 |
7,010 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | 3 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | Let $\theta$ be the smaller angle in the triangle. Then the sum of shorter and longer leg is $\sqrt{441}(2+\tan{\theta}+\cot{\theta})$ . We observe that the short leg has length $\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})$ . Grouping and squaring, we get $\sqrt{\frac{440}{441}} = \frac{\sin{\the... | 462 |
7,011 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_15 | 4 | Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$ , as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$
AIME 1987 Problem 15.png | [asy] size(200); pair A, B, C, D, E, F; A = (0, 5); B = (12, 0); C = (0, 0); D = (0, 60/17); E = (60/17, 60/17); F = (60/17, 0); draw(A--B--C--cycle); draw(D--E--F); label("$A$",A,N); label("$B$",B,dir(0)); label("$C$",C,SW); label("$D$",D,W); label("$E$",E,NE); label("$F$",F,S); label("$S_1$",(30/17,30/17)); [/asy]
[a... | 462 |
7,012 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_1 | 1 | What is the sum of the solutions to the equation $\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$ | Let $y = \sqrt[4]{x}$ . Then we have $y(7 - y) = 12$ , or, by simplifying, \[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\]
This means that $\sqrt[4]{x} = y = 3$ or $4$
Thus the sum of the possible solutions for $x$ is $4^4 + 3^4 = \boxed{337}$ | 337 |
7,013 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | 1 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | More generally, let $(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)$ so that $\left(x^2,y^2,z^2\right)=(5,6,7).$
We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x... | 104 |
7,014 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | 2 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{4... | 104 |
7,015 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 | 3 | Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | Notice that in a triangle with side-lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7,$ by Heron's Formula, the area is the square root of the original expression.
Let $\theta$ be the measure of the angle opposite the $2\sqrt7$ side. By the Law of Cosines, \[\cos\theta=\frac{\left(2\sqrt5\right)^2+\left(2\sqrt{6}\right)^2-\left(... | 104 |
7,016 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | 1 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | Since $\cot$ is the reciprocal function of $\tan$
$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$
Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$
Using the tangent addition formula:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\t... | 150 |
7,017 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | 2 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | Using the formula for tangent of a sum, $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}$ . We only need to find $\tan x \tan y$
We know that $25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$ . Cross multiplying, we have $\frac{\sin x \cos y + \cos x \sin y}{\cos x \c... | 150 |
7,018 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_3 | 3 | If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$ | Let $a=\tan x$ and $b=\tan y$ . This simplifies the equations to:
\[a + b = 25\]
\[\frac{1}{a} + \frac{1}{b} = 30\]
Taking the tangent of a sum formula from Solution 2, we get $\tan(x+y) = \frac{25}{1 - ab}$
We can use substitution to solve the system of equations from above: $b = -a + 25$ , so $\frac{1}{a} + \frac{1}{... | 150 |
7,019 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_4 | 1 | Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below. | Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ . Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$ , so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$ | 181 |
7,020 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_4 | 2 | Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below. | Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, \[3x_4... | 181 |
7,021 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | 1 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $... | 981 |
7,022 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | 2 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | Notice that the first term of the sequence is $1$ , the second is $3$ , the fourth is $9$ , and so on. Thus the $64th$ term of the sequence is $729$ . Now out of $64$ terms which are of the form $729$ $'''S'''$ $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$ , i.e. $972$ , is greater t... | 981 |
7,023 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | 3 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | After the $n$ th power of 3 in the sequence, the number of terms after that power but before the $(n+1)$ th power of 3 is equal to the number of terms before the $n$ th power, because those terms after the $n$ th power are just the $n$ th power plus all the distinct combinations of powers of 3 before it, which is just ... | 981 |
7,024 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | 4 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the $2^{nth}$ term is equal to $3^n$ . From here, we can ballpark the range of the 100th term. The 64th term is $3^6$ $729$ and the 128th term is $3^7$ $2187$ . Writing out more terms of the sequence until the next power of 3 a... | 981 |
7,025 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_7 | 5 | The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | The number of terms $3^n$ produces includes each power of 3 ( $1, 3^1, ..., 3^n$ ), the sums of two power of 3s(ex. $3^1 + 1$ ), three power of 3s (ex. $3^1 + 1 + 3^n$ ), all the way to the sum of them all. Since there are $n+1$ powers of 3, the one number sum gives us ${n+1\choose 1}$ terms, the two number ${n+1\choos... | 981 |
7,026 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | 1 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | The prime factorization of $1000000 = 2^65^6$ , so there are $(6 + 1)(6 + 1) = 49$ divisors, of which $48$ are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.
Writing out the first few terms, we see that the answer is equal to \[\log 1 + \log 2 + \log 4 ... | 141 |
7,027 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | 2 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | The formula for the product of the divisors of $n$ is $n^{(d(n))/2}$ , where $d(n)$ is the number of divisor of $n$ .
We know that $\log_{10} a + \log_{10} b + \log_{10} c + \log_{10} d...$ and so on equals $\log_{10} (abcd...)$ by sum-product rule of logs, so the problem is reduced to finding the logarithm base 10 of ... | 141 |
7,028 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | 3 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | Since the prime factorization of $10^6$ is $2^6 \cdot 5^6$ , the number of factors in $10^6$ is $7 \cdot 7=49$ . You can pair them up into groups of two so each group multiplies to $10^6$ . Note that $\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6$ . Thus, the sum of the logs of the divisors is half the number of d... | 141 |
7,029 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_8 | 4 | Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$ . What is the integer nearest to $S$ | Note that we can just pair terms up such that the product is $10^{6}.$ Now, however, note that $10^{3}$ is not included. Therefore we first exclude. We have $\displaystyle\frac{49-1}{2} = 24$ pairs that all multiply to $10^{6}.$ Now we include $10^{3}$ so our current product is $24 \cdot 6 - 3.$ However we dont want to... | 141 |
7,030 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | 1 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are p... | 306 |
7,031 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | 2 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are p... | 306 |
7,032 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | 3 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Refer to the diagram in solution 2; let $a^2=[E'EP]$ $b^2=[D'DP]$ , and $c^2=[F'FP]$ . Now, note that $[E'BD]$ $[D'DP]$ , and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$ . Similarly... | 306 |
7,033 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_9 | 4 | In $\triangle ABC$ $AB= 425$ $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle . If these three segments are of an equal length $d$ , find $d$ | Refer to the diagram above. Notice that because $CE'PF$ $AF'PD$ , and $BD'PE$ are parallelograms, $\overline{DD'} = 425-d$ $\overline{EE'} = 450-d$ , and $\overline{FF'} = 510-d$
Let $F'P = x$ . Then, because $\triangle ABC \sim \triangle F'PF$ $\frac{AB}{AC}=\frac{F'P}{F'F}$ , so $\frac{425}{510}=\frac{x}{510-d}$ . Si... | 306 |
7,034 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | 1 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reve... | Let $m$ be the number $100a+10b+c$ . Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$ . But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$ .
Recall that $a, b, c$ refer to the digits the three digit number $(abc)$... | 358 |
7,035 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | 2 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reve... | As in Solution 1, $3194 + m \equiv 222(a+b+c) \pmod{222}$ , and so as above we get $m \equiv 136 \pmod{222}$ .
We can also take this equation modulo $9$ ; note that $m \equiv a+b+c \pmod{9}$ , so
\[3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.\]
Therefore $m$ is $7$ mod $9$ and $136$ mod $222$... | 358 |
7,036 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | 3 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reve... | Let $n=abc$ then \[N=222(a+b+c)-n\] \[N=222(a+b+c)-100a-10b-c=3194\] Since $0<100a+10b+c<1000$ , we get the inequality \[N<222(a+b+c)<N+1000\] \[3194<222(a+b+c)<4194\] \[14<a+b+c<19\] Checking each of the multiples of $222$ from $15\cdot222$ to $18\cdot222$ by subtracting $N$ from each $222(a+b+c)$ , we quickly find $n... | 358 |
7,037 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_10 | 4 | In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reve... | The sum of the five numbers is $222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194$ We can see that $3194 \equiv 8$ （mod $9$ ） and $122 \equiv 5$ （mod $9$ ） so we need to make sure that $a+b+c \equiv 7$ （mod $9$ ） by some testing. So we let $a+b+c=9k+7$
Then, we know that $1\leq a+b+c \leq 27$ so only $7,1... | 358 |
7,038 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | 1 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in ... | 816 |
7,039 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | 2 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Again, notice $x = y - 1$ . So
\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.
We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem... | 816 |
7,040 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | 3 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Again, notice $x=y-1$ . Substituting $y-1$ for $x$ in $f(x)$ gives: \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. From binomial theorem, the coefficient of the $y^2$ term is ${2\ch... | 816 |
7,041 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | 4 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Let $f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$
Then, since $f(x)=g(y)$ \[\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}\] $\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}$ by the power rule.
Similarly, $\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\c... | 816 |
7,042 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_11 | 5 | The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$ | Let $V$ be the vector space of polynomials of degree $\leq 17,$ and let $B = \{1, x, x^2, ..., x^{17} \}$ and $C = \{1, (x+1), (x+1)^2, ..., (x+1)^{17} \}$ be two bases for $V$ .
Let $\vec{v} \in V$ be the polynomial given in the problem, and it is easy to see that $[ \vec{v} ]_B = \langle 1, -1, 1, -1, ... , 1, -1 \ra... | 816 |
7,043 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_13 | 1 | In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH HH , and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH , three HT , four TH ... | Let's consider each of the sequences of two coin tosses as an operation instead; this operation takes a string and adds the next coin toss on (eg, THHTH HT THHTHT ). We examine what happens to the last coin toss. Adding HH or TT is simply an identity for the last coin toss, so we will ignore them for now. However, addi... | 560 |
7,044 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_14 | 1 | The shortest distances between an interior diagonal of a rectangular parallelepiped $P$ , and the edges it does not meet are $2\sqrt{5}$ $\frac{30}{\sqrt{13}}$ , and $\frac{15}{\sqrt{10}}$ . Determine the volume of $P$ | In the above diagram, we focus on the line that appears closest and is parallel to $BC$ . All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$ , the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and ... | 750 |
7,045 | https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_15 | 1 | Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$ . Given that the length of the hypotenuse $AB$ is $60$ , and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$ | We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$ [asy] import graph; size(150); Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 2.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return (x + 3); } real g( real x){ return (2x + 4); } draw(graph(f,-8,5),red+linewidth(1)); draw(graph(g... | 400 |
7,046 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_1 | 2 | Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$ | Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
\[x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{... | 384 |
7,047 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_2 | 2 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | Let $a$ $b$ be the $2$ legs, we have the $2$ equations \[\frac{a^2b\pi}{3}=800\pi,\frac{ab^2\pi}{3}=1920\pi\] Thus $a^2b=2400, ab^2=5760$ . Multiplying gets \begin{align*} (a^2b)(ab^2)&=2400\cdot5760\\ (ab)^3&=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5)\\ ab&=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \end{align*} Adding gets \be... | 26 |
7,048 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_3 | 1 | Find $c$ if $a$ $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$ | Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$ . Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$ . Since $a, b$ are integers , this means $b$ is a divisor of 10... | 198 |
7,049 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_4 | 1 | A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png | The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram . Using the smaller side of the parallelogram, $1/n$ , as the base, where the height is 1, we find that the area of the parallelogram is $A = \frac{1}{n}$ . By the Pythagorean Theorem , the longer base of... | 32 |
7,050 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_4 | 2 | A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png | Aime.png
Surrounding the square with area $\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \frac{1}{1985} = 1$ , where $X$ is the area of the of the 4 triangles.
We can thus use proportions to solve this problem. \begin{eqnarray*} \frac{GF}{BE}=\frac{CG}{CB}\implies \... | 32 |
7,051 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_4 | 3 | A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png | AIME 1985 Problem 4 Solution 3 Diagram.png
Line Segment $DE = \frac{1}{n}$ , so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$ . Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$ , as it is the same length as the sides of the square. Notice that $\triangle C... | 32 |
7,052 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | 1 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | Let the interior point be $P$ , let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Let $x$ be the area of $\triangle APE$ and $y$ be the area of $\triangle CPD$ . Note that $\triangle APF$ and $\triangle BPF$ share the same altitude from $P$ , so the ratio of thei... | 315 |
7,053 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | 2 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $\times$ side) = (Other weight) $\times$ (The other side), the problem yi... | 315 |
7,054 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_6 | 3 | As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$
AIME 1985 Problem 6.png | Let the interior point be $P$ and let the points on $\overline{BC}$ $\overline{CA}$ and $\overline{AB}$ be $D$ $E$ and $F$ , respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $\frac{FB\cdot DC\cdot EA}{DB\cdot CE\cdot AF}=1.$ However, we can deduce $\frac{FB}{AF}=\frac{3}{4}$ from the fact that $[A... | 315 |
7,055 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_7 | 1 | Assume that $a$ $b$ $c$ , and $d$ are positive integers such that $a^5 = b^4$ $c^3 = d^2$ , and $c - a = 19$ . Determine $d - b$ | It follows from the givens that $a$ is a perfect fourth power $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube . Thus, there exist integers $s$ and $t$ such that $a = t^4$ $b = t^5$ $c = s^2$ and $d = s^3$ . So $s^2 - t^4 = 19$ . We can factor the left-hand side of this equation as a d... | 757 |
7,056 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | 1 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Noting that all of the numbers are even, we can reduce this to any real number $x$ between $0$ to $\frac 12$ , as this will be equivalent to $\frac n2$ to $\frac {n+1}2$ for any integer $n$ (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers w... | 600 |
7,057 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | 3 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Because $2,4,6,8$ are all multiples of $2$ , we can speed things up. We only need to check up to $\frac{12}{24}$ , and the rest should repeat. As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{12}{24}$ . Similarly, this should repeat 100 times, for $\boxed{600}$ | 600 |
7,058 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | 4 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | We only need to check the numbers where it increments, namely $\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}$ . As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{1}{2}$ . Similarly, this should repeat 100 times, for $\boxed{600}$ | 600 |
7,059 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | 5 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Recall from Hermite's Identity that $\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor$ . Then we can rewrite $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloo... | 600 |
7,060 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | 6 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | Let $x=\lfloor x\rfloor+\{x\}$ then \begin{align*} \lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\ &=2\lfloor x\rfloor+4\lfloor x... | 600 |
7,061 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_10 | 7 | How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$ | To simplify the question, let $y = 2x$ . Then, the expression in the question becomes $\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$
Let $\{x\}$ represent the non-integer part of $x$ (For example, $\{2.8\} = 0.8$ ). Then,
\begin{align*} \lfloor y \rfloor + \lfloor 2y \rfloor + \lfloo... | 600 |
7,062 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 1 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | We evaluate $P(7)$ recursively: \begin{alignat*}{6} P(0)&=1, \\ P(1)&=\frac13(1-P(0))&&=0, \\ P(2)&=\frac13(1-P(1))&&=\frac13, \\ P(3)&=\frac13(1-P(2))&&=\frac29, \\ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\ P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\ P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\ P(7)&=\frac13(1-P(6))&&=\frac... | 182 |
7,063 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 2 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Let $P(k)=Q(k)+c$ for some function $Q(k)$ and constant $c.$ For all $k\geq1,$ the recursive formula for $P(k)$ becomes \[Q(k)+c=\frac13(1-(Q(k-1)+c))=\frac13-\frac13Q(k-1)-\frac13c.\] Solving for $Q(k),$ we get \[Q(k)=\frac13-\frac13Q(k-1)-\frac43c.\] For simplicity purposes, we set $c=\frac14,$ which gives \[Q(k)=-\f... | 182 |
7,064 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 3 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Denominator
There are $3^7$ ways for the bug to make $7$ independent crawls without restrictions.
Numerator
Let $V_k$ denote the number of ways for the bug to crawl exactly $k$ meters starting from vertex $V$ and ending at vertex $A,$ where $V\in\{A,B,C,D\}$ and $k$ is a positive integer. We wish to find $A_7.$
Since t... | 182 |
7,065 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 4 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Define notation $V_k$ as Solution 2 does.
In fact, we can generalize the following relationships for all nonnegative integers $k:$ \begin{align*} A_0&=1, \\ B_0&=0, \\ C_0&=0, \\ D_0&=0, \\ A_{k+1}&=B_k+C_k+D_k, \\ B_{k+1}&=A_k+C_k+D_k, \\ C_{k+1}&=A_k+B_k+D_k, \\ D_{k+1}&=A_k+B_k+C_k. \\ \end{align*} Using these equat... | 182 |
7,066 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 5 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Let $a_n$ denotes the number of ways that the bug arrives at $A$ after crawling $n$ meters, then we have $a_1=0$
Notice that there is respectively $1$ way to arrive at $A$ for each of the different routes after the previous $n-1$ crawls, excluding the possibility that the bug ends up at $A$ after the $(n-1)$ th crawl (... | 182 |
7,067 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 6 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Let $A(n)$ be the probability the bug lands on vertex $A$ after crawling $n$ meters, $B(n)$ be the probability the bug lands on vertex $B$ after crawling $n$ meters, and etc.
Note that $A(1)=0$ and $B(1)=C(1)=D(1)=\frac13.$ For $n\geq2,$ the probability that the bug land on each vertex after $n$ meters is $\frac13$ the... | 182 |
7,068 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 7 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Let $\omega = e^{i\pi / 2}$ . We have that if $G(x) = (x+x^2+x^3)^7$ , then \[\frac{G(1) + G(\omega) + G(\omega^2) + G(\omega^3)}{4} = \frac{2187-1-1-1}{4} = 546.\] From here, the desired probability is $\frac{546}{2187} = \frac{182}{729}$ . Therefore, the answer is $n=\boxed{182}$ | 182 |
7,069 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 8 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | We can factor $(x+x^2+x^3)^7$ as $x^7(1+x+x^2)^7.$ The $x^{4n}$ coefficients of $(x+x^2+x^3)^7$ will be the same as the $x^{4n+1}$ coefficients of $(1+x+x^2)^7.$ The possible values for $4n+1$ would then be $1,$ $5,$ $9,$ and $13.$ Because $1+13=5+9=14,$ the coefficients of $x^1$ and $x^{13}$ are equal and so are the c... | 182 |
7,070 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 9 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | We can find the number of different times the bug reaches vertex $A$ before the $7$ th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A.$
Define $f(x)$ to be the number of paths of length $x$ which start and end at $A$ but do not pass through $A$ otherwise. Obvi... | 182 |
7,071 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 10 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | Note that this problem is basically equivalent to the following:
How many distinct sequences of $8$ integers $a_1, a_2, a_3, \ldots, a_8$ are there such that $a_1 = a_8 = 1,$ $a_i \in \{1, 2, 3, 4\}$ for all $2 \leq i\leq 8,$ and $a_i \neq a_{i+1}$ for all $1 \leq i \leq 7$
Now consider the $8$ integers modulo $4.$ Let... | 182 |
7,072 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 | 11 | Let $A$ $B$ $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$ , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the... | We instead find the probability that the bug is NOT at vertex $A$ after crawling $n$ meters (equivalent to moving $n$ times). Call $A_n$ the probability that the bug IS at vertex $A$ after $n$ moves; call $O_n$ the probability that the bug is on some other vertex. We have the following recurrence relations. \[A_n = \fr... | 182 |
7,073 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_13 | 2 | The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers | We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm
$a_{n+1}-a_n = 2n+1$
Now, the question is to find the GCD of $2n+1$ and $100+n^2$ .
We subtract $2n+1$ $100$ times from $100+n^2$ \[(100+n^2)-100(2n+1)\] \... | 401 |
7,074 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | 1 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the poin... | Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned. By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10. Now, the 10 weakes... | 25 |
7,075 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | 2 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the poin... | Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people... | 25 |
7,076 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_14 | 3 | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the poin... | Note that the total number of points accumulated must sum to ${p \choose 2} = \frac{p(p-1)}{2}$ . Say the number of people is $n$ . Consider the number of points gained when the 10 lowest scoring people play each other. The problem tells us that each of these 10 people must earn exactly half of the total number of poin... | 25 |
7,077 | https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_15 | 1 | Three 12 cm $\times$ 12 cm squares are each cut into two pieces $A$ and $B$ , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon , as shown in the second figure, so as to fold into a polyhedron . What is the volume (in $\mathrm{cm}... | Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\circ$ from each other) yields a cube , so the volume is $\frac12 \cdot 12^3 = 864$ , so our answer is $\boxed{864}$ | 864 |
7,078 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_1 | 2 | Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$ | We want to find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ , which can be rewritten as $a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}$ .
We can split $a_1+a_2+a_3+\ldots+a_{98}$ into two parts: \[a_1+a_2+a_3+\ldots+a_{49}\] and \[a_{50}+a_{51}+a_{52}+\ldots+a_{98}\] Note th... | 93 |
7,079 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_2 | 1 | The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$ | Any multiple of 15 is a multiple of 5 and a multiple of 3.
Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$ . For thi... | 592 |
7,080 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3 | 1 | point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ | By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \dfrac{ab\sin C}{2}$ to s... | 144 |
7,081 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3 | 2 | point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ | Alternatively, since the triangles are similar by $AA$ , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\dfrac{base}{height} = \dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, th... | 144 |
7,082 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3 | 3 | point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ | The base of $\triangle{ABC}$ is $BC$ . Let the base of $t_1$ be $x$ , the base of $t_2$ be $y$ , and the base of $t_3$ be $z$ . Since $\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$ . We can relate $t_1,t_2... | 144 |
7,083 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_3 | 4 | point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ | Since the three lines through $P$ are parallel to the sides, $t_1$ $t_2$ $t_3$ , and $\triangle{ABC}$ are similar by $AA$ similarity. Suppose the area of $\triangle{ABC}$ is $x^2$ , so the ratio of the base of $t_1$ to the base of $t_2$ to the base of $t_3$ to the base of $\triangle{ABC}$ is $2:3:7:x$ . Because the qua... | 144 |
7,084 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_4 | 1 | Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$ . However, if $68$ is removed, the average of the remaining numbers drops to $55$ . What is the largest number that can appear in $S$ | Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $... | 649 |
7,085 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_4 | 2 | Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$ . However, if $68$ is removed, the average of the remaining numbers drops to $55$ . What is the largest number that can appear in $S$ | Suppose that $S$ has $n$ numbers other than the $68.$ We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ \hline & & & \\ [-2.5ex] \textbf{Initial} & n+1 & 56 & 56(n+1) \\ \hline & & & \\ [-2.5ex] \textbf{Final} & n & 55 & 55n \end{array}... | 649 |
7,086 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5 | 1 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ | Use the change of base formula to see that $\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5$ ; combine denominators to find that $\frac{\log ab^3}{3\log 2} = 5$ . Doing the same thing with the second equation yields that $\frac{\log a^3b}{3\log 2} = 7$ . This means that $\log ab^3 = 15\log 2 \Longrightarrow ab^3 = ... | 512 |
7,087 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5 | 2 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ | We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become $\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5$ and $\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7$ . Adding the equations and factoring, we get $(\frac{1}{\ln 8}+\... | 512 |
7,088 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5 | 3 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ | This solution is very similar to the above two, but it utilizes the well-known fact that $\log_{m^k}{n^k}= \log_m{n}.$ Thus, $\log_8a+\log_4b^2=5 \Rightarrow \log_{2^3}{(\sqrt[3]{a})^3} + \log_{2^2}{b^2} = 5 \Rightarrow \log_2{\sqrt[3]{a}} + \log_2{b} = 5 \Rightarrow \log_2{\sqrt[3]{a}b} = 5.$ Similarly, $\log_8b+\log... | 512 |
7,089 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5 | 4 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ | We can change everything to a common base, like so: $\log_8{a} + \log_8{b^3} = 5,$ $\log_8{b} + \log_8{a^3} = 7.$ We set the value of $\log_8{a}$ to $x$ , and the value of $\log_8{b}$ to $y.$ Now we have a system of linear equations: \[x + 3y = 5,\] \[y + 3x = 7.\] Now add the two equations together then simplify, we'l... | 512 |
7,090 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5 | 5 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ | Add the two equations to get $\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12$ . This can be simplified with the log property $\log_n {x}+\log_n {y}=log_n {xy}$ . Using this, we get $\log_8 {ab}+ \log_4 {a^2b^2}=12$ . Now let $\log_8 {ab}=c$ and $\log_4 {a^2b^2}=k$ . Converting to exponents, we get $8^c=ab$ and $4^k=(... | 512 |
7,091 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_5 | 6 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ | By properties of logarithms, we know that $\log_8 {a}+ \log_4 {b ^ 2} = \log_2 {a ^ {1/3}}+ \log_2 {b} = 5$
Using the fact that $\log_a {b} + \log_a {c} = log_a{b*c}$ , we get $\log_2 {a^{1/3} * b} = 5$
Similarly, we know that $\log_2 {a * b^{1/3}} = 7$
From these two equations, we get $a^{1/3} * b = 2^5$ and $a * b^... | 512 |
7,092 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_6 | 2 | Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the ... | First of all, we can translate everything downwards by $76$ and to the left by $14$ . Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem:
Two circles, each of radius $3$ , are drawn with centers... | 24 |
7,093 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_7 | 1 | The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$
Find $f(84)$ | Define $f^{h} = f(f(\cdots f(f(x))\cdots))$ , where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \ldots f^{y}(1004)$ $1004 = 84 + 5(y - 1) \Longrightarrow y = 185$ . So we now need to reduce $f^{185}(1004)$
Let’s write out a couple more iterations of this function: \begi... | 997 |
7,094 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_7 | 2 | The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$
Find $f(84)$ | Assume that $f(84)$ is to be performed $n+1$ times. Then we have \[f(84)=f^{n+1}(84)=f(f^n(84+5))\] In order to find $f(84)$ , we want to know the smallest value of \[f^n(84+5)\ge1000\] Because then \[f(84)=f(f^n(84+5))=(f^n(84+5))-3\] From which we'll get a numerical value for $f(84)$
Notice that the value of $n$ we e... | 997 |
7,095 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_8 | 1 | The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane . Determine the degree measure of $\theta$ | We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$ , then $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\circ$ for integer $m$ (the ninth degree roots of unity ). Now we simply need to... | 160 |
7,096 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_8 | 2 | The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane . Determine the degree measure of $\theta$ | The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$ . Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$ , which have arguments of $120$ and $240,$ respectively.
We can write them as $z^3 = \cos 240^\circ + i\sin 240^\circ$ and $z^3 = \cos 120^\circ + i\sin 120^\circ$ .
S... | 160 |
7,097 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_9 | 3 | In tetrahedron $ABCD$ edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$ | We can use 3D coordinates.
Let $A = (0, 0, 0)$ and $B = (3, 0, 0).$ WLOG, let $D = \left(\frac{3}{2}, 8, 0\right)$ , because the area of $\Delta{ABD} = 12$ and the tetrahedron area won't change if we put it somewhere else with $y=8.$
To find $C$ , we can again let the $x$ -coordinate be $\frac{3}{2}$ for simplicity. No... | 20 |
7,098 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10 | 1 | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall th... | Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then \begin{align*} s&=30+4c-w \\ &=30+4(c-1)-(w-4) \\ &=30+4(c+1)-(w+4). \end{align*} Therefore, Mary could not have left at least five blank; otherwise, one more correct and four more wrong would produce the same score. Similarly, Mary could ... | 119 |
7,099 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10 | 2 | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall th... | A less technical approach that still gets the job done:
Pretend that the question is instead a game, where we are trying to get certain numbers by either adding $4$ or $5.$ The maximum number we can get is $70.$ The goal of the game is to find out what number we can achieve in only ONE method, while all other numbers a... | 119 |
7,100 | https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10 | 3 | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall th... | Based on the value of $c,$ we construct the following table: \[\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} &\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&\hspace{5.5mm}&&&&&&&&&&&&& \\ [-2.5ex] \boldsymbol{c} &\boldsymbol{\cdots}&\boldsymbol{12}&\boldsymbol{1... | 119 |
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