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int64 1
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6,301
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_3
| 2
|
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.
|
Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits.
Let $N=\overline{abc}$ , so $N=100a+10b+c$ . Thus, $N'=\overline{bc}=10b+c$ . By the constraints of the problem, we see that $N=29N'$ , so \[100a+10b+c=29(10b+c).\] Now, we subtract and divide to get \[100a=28(10b+c)\] \[25a=70b+7c.\] Clearly, $c$ must be a multiple of $5$ because both $25a$ and $70b$ are multiples of $5$ . Thus, $c=5$ . Now, we plug that into the equation: \[25a=70b+7(5)\] \[25a=70b+35\] \[5a=14b+7.\] By the same line of reasoning as earlier, $a=7$ . We again plug that into the equation to get \[35=14b+7\] \[b=2.\] Now, since $a=7$ $b=2$ , and $c=5$ , our number $N=100a+10b+c=\boxed{725}$
| 725
|
6,302
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_4
| 1
|
Let $N$ be the number of consecutive $0$ 's at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by $1000$
|
A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression. Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
One way to do this is as follows: $96$ of the numbers $1!,\ 2!,\ 3!,\ 100!$ have a factor of $5$ $91$ have a factor of $10$ $86$ have a factor of $15$ . And so on. This gives us an initial count of $96 + 91 + 86 + \ldots + 1$ . Summing this arithmetic series of $20$ terms, we get $970$ . However, we have neglected some powers of $5$ - every $n!$ term for $n\geq25$ has an additional power of $5$ dividing it, for $76$ extra; every n! for $n\geq 50$ has one more in addition to that, for a total of $51$ extra; and similarly there are $26$ extra from those larger than $75$ and $1$ extra from $100$ . Thus, our final total is $970 + 76 + 51 + 26 + 1 = 1124$ , and the answer is $\boxed{124}$
| 124
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6,303
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5
| 1
|
The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$
|
We begin by equating the two expressions:
\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]
Squaring both sides yields:
\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]
Since $a$ $b$ , and $c$ are integers, we can match coefficients:
\begin{align*} 2ab\sqrt{6} &= 104\sqrt{6} \\ 2ac\sqrt{10} &=468\sqrt{10} \\ 2bc\sqrt{15} &=144\sqrt{15}\\ 2a^2 + 3b^2 + 5c^2 &=2006 \end{align*}
Solving the first three equations gives: \begin{eqnarray*}ab &=& 52\\ ac &=& 234\\ bc &=& 72 \end{eqnarray*}
Multiplying these equations gives $(abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}$
| 936
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6,304
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_5
| 2
|
The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers . Find $abc$
|
We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting $x=\sqrt{2}$ $y=\sqrt{3}$ , and $z=\sqrt{5}$ . Since
\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]
we attempt to rewrite the radicand in this form:
\[2006+2(52xy+234xz+72yz)\]
Factoring, we see that $52=13\cdot4$ $234=13\cdot18$ , and $72=4\cdot18$ . Setting $p=13$ $q=4$ , and $r=18$ , we see that
\[2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5\]
so our numbers check. Thus $104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2$ . Square rooting gives us $13\sqrt{2}+4\sqrt{3}+18\sqrt{5}$ and our answer is $13\cdot4\cdot18=\boxed{936}$
| 936
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6,305
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6
| 1
|
Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$
|
Numbers of the form $0.\overline{abc}$ can be written as $\frac{abc}{999}$ . There are $10\times9\times8=720$ such numbers. Each digit will appear in each place value $\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\frac{45\times72\times111}{999}= \boxed{360}$
| 360
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6,306
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_6
| 2
|
Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits . Find the sum of the elements of $\mathcal{S}.$
|
Alternatively, for every number, $0.\overline{abc}$ , there will be exactly one other number, such that when they are added together, the sum is $0.\overline{999}$ , or, more precisely, 1. As an example, $.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1$
Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is $\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}$
| 360
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6,307
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7
| 1
|
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]
|
Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof
Let the set of parallel lines be perpendicular to the x-axis , such that they cross it at $0, 1, 2 \ldots$ . The base of region $\mathcal{A}$ is on the line $x = 1$ . The bigger base of region $\mathcal{D}$ is on the line $x = 7$ .
Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem.
Since the area of the triangle is equal to $\frac{1}{2}bh$
\[\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}\]
Solve this to find that $s = \frac{5}{6}$
Using the same reasoning as above, we get $\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$ , which is $\boxed{408}$
| 408
|
6,308
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_7
| 3
|
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]
|
Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.$ Therefore the ratio we seek is $\frac{66(ay)}{11xy} =\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \implies ab = 6bx \implies a = 6x$ so the final ratio is $6 \cdot 68 = \boxed{408}.$
| 408
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6,309
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_11
| 1
|
A collection of 8 cubes consists of one cube with edge length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:
Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ is divided by 1000?
|
We proceed recursively . Suppose we can build $T_m$ towers using blocks of size $1, 2, \ldots, m$ . How many towers can we build using blocks of size $1, 2, \ldots, m, m + 1$ ? If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \ldots, m$ . Given a tower using blocks $1, 2, \ldots, m$ (with $m \geq 2$ ), we can insert the block of size $m + 1$ in exactly 3 places: at the beginning, immediately following the block of size $m - 1$ or immediately following the block of size $m$ . Thus, there are 3 times as many towers using blocks of size $1, 2, \ldots, m, m + 1$ as there are towers using only $1, 2, \ldots, m$ . There are 2 towers which use blocks $1, 2$ , so there are $2\cdot 3^6 = 1458$ towers using blocks $1, 2, \ldots, 8$ , so the answer is $\boxed{458}$
| 458
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6,310
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_12
| 1
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Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$ , where $x$ is measured in degrees and $100< x< 200.$
|
Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$ , we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$ . But $a+b = 2\cos 4x\cos x$ , so we require $\cos x = 0$ $\cos 3x = 0$ $\cos 4x = 0$ , or $\cos 5x = 0$
Hence we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$
| 906
|
6,311
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13
| 1
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For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square
|
Given $g : x \mapsto \max_{j : 2^j | x} 2^j$ , consider $S_n = g(2) + \cdots + g(2^n)$ . Define $S = \{2, 4, \ldots, 2^n\}$ . There are $2^0$ elements of $S$ that are divisible by $2^n$ $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ elements of $S$ that are divisible by $2^1$ but not by $2^2$
Thus \begin{align*} S_n &= 2^0\cdot2^n + 2^0\cdot2^{n-1} + 2^1\cdot2^{n-2} + \cdots + 2^{n-2}\cdot2^1\\ &= 2^n + (n-1)2^{n-1}\\ &= 2^{n-1}(n+1).\end{align*} Let $2^k$ be the highest power of $2$ that divides $n+1$ . Thus by the above formula, the highest power of $2$ that divides $S_n$ is $2^{k+n-1}$ . For $S_n$ to be a perfect square, $k+n-1$ must be even. If $k$ is odd, then $n+1$ is even, hence $k+n-1$ is odd, and $S_n$ cannot be a perfect square. Hence $k$ must be even. In particular, as $n<1000$ , we have five choices for $k$ , namely $k=0,2,4,6,8$
If $k=0$ , then $n+1$ is odd, so $k+n-1$ is odd, hence the largest power of $2$ dividing $S_n$ has an odd exponent, so $S_n$ is not a perfect square.
In the other cases, note that $k+n-1$ is even, so the highest power of $2$ dividing $S_n$ will be a perfect square. In particular, $S_n$ will be a perfect square if and only if $(n+1)/2^{k}$ is an odd perfect square.
If $k=2$ , then $n<1000$ implies that $\frac{n+1}{4} \le 250$ , so we have $n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2$
If $k=4$ , then $n<1000$ implies that $\frac{n+1}{16} \le 62$ , so $n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2$
If $k=6$ , then $n<1000$ implies that $\frac{n+1}{64}\le 15$ , so $n+1=64,64\cdot 3^2$
If $k=8$ , then $n<1000$ implies that $\frac{n+1}{256}\le 3$ , so $n+1=256$
Comparing the largest term in each case, we find that the maximum possible $n$ such that $S_n$ is a perfect square is $4\cdot 3^2 \cdot 5^2 - 1 = \boxed{899}$
| 899
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6,312
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13
| 2
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For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square
|
First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even.
so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$ . However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$
| 899
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6,313
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13
| 3
|
For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square
|
At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations.
We first simplify all the messy notation in the $S_n$ term. Note that the problem asks us to find the smallest value of $n<1000$ such that there exists an integer $k$ that satisfies
\[k^2 = g(2) + g(4) + \cdots + g(2^n)\]
Since there is no obvious way to approach this problem, we start by experimenting with small values of $n$ to evaluate some $S_n$
We play with these values:
\[S_1 = g(2) = 2\]
\[S_2 = g(2) + g(4) = 2+4 = 6\]
\[S_3 = g(2) + g(4) + g(6) + g(8) = 16\]
\[S_4 = g(2) + g(4) + g(6) + g(8) + g(10) +g(12)+g(14)+g(16) = 40\]
We are certainly not going to expand all of this out... so let's look for patterns from these $4$ values!
Using a little bit of ingenuity, we note that
\[S_2 = 2+4 = S_1 + 4\]
\[S_3 = 2+4+2+8 = 8+8 = S_2 + S_1 + 8\]
\[S_4 = 2+4+2+8+2+4+2+16 = S_3 + S_2 + S_1 + 16\]
Aha! We see powers of two in each of our terms! Therefore, we can say that
\[S_2 = S_1 + 2^2\]
\[S_3 = S_2+S_1 + 2^3\]
\[S_4 = S_3 + S_2 + S_1 + 2^4\]
Hooray! We have a recursion! Realistically, we would want to prove that the recursion works, but I currently don't know how to prove it.
On the actual AIME, go with whatever patterns you see, because most likely those are the patterns that the test-makers want the students to see.
So we may generalize a formula for $S_n$
\[S_n = 2^n + S_{n-1} + S_{n-2} + \cdots + S_2 + S_1\]
Uh oh... this formula is not in closed form. Looks like we'll have to use our recursion to develop one manually. We do so by using our recursion for $S_{n-1}$
\[S_n = 2^n + (2^{n-1} + S_{n-2} + S_{n-3} + \cdots + S_2 + S_1) + S_{n-2} + S_{n-3} + \cdots + S_2 + S_1\]
\[S_n = 2^n + 2^{n-1} + 2 (S_{n-2} + S_{n-3} + \cdots + S_2 + S_1\]
Let's simplify a bit further, where we use our recursion for $S_{n-2}$
\[S_n = 2^n + 2^{n-1} +(2S_{n-2}) + 2(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1)\]
\[S_n = 2^n + 2^{n-1} + 2(2^{n-2}) + 2(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1) + 2(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1)\]
\[S_n = 2^n + 2^{n-1} + 2^{n-1} + 4(S_{n-3} + S_{n-4} + \cdots + S_2 + S_1)\]
We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form:
\[S_n = 2^n + \underbrace{2^{n-1}}_{n-2} + 2^{n-2}(S_1)\]
\[S_n = 2^n + 2^{n-1}(n-2) + 2^{n-1}\]
\[S_n = 2^n + 2^{n-1}(n-1)\]
\[S_n = 2^{n-1}(2 + (n-1)\]
\[S_n = 2^{n-1}(n+1)\]
We have our closed form, so now we can find the largest value of $n$ such that $S_n$ is a perfect square.
In order for $S_n$ to be a perfect square, we must have $n-1$ even and $n+1$ be a perfect square.
Since $n<1000$ , we have $n+1 < 1001$ . We first try $n+1 = 31^2 = 961$ (since it is the smallest square below $1000$ , which gives us $n=960$ . But $n-1$ isn't even, so we discard this value.
Next, we try the second smallest value, which is $n = 30^2 = 900$ , which tells us that $n=899$ $n-1$ is indeed even, and $n+1$ is a perfect square, so the largest value of $n$ such that $S_n$ is a perfect square is $899$
Our answer is $\boxed{899}$
| 899
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6,314
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_13
| 4
|
For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square
|
First, we set intervals. Say that a number $N$ falls strictly within $[2^k, 2^{k+1}]$
$N<2^{k+1}=2^k+2^k$
We can generalize this:
If a number is in the form $N=2^k+2^{R}O$ where $O$ is a positive odd number, $R<k$
$N<2^{k+1}=2^k+2^k\Longrightarrow O<2^{k-R}$ so there are $2^{k-R-1}$ numbers that satisfy this form.
For all numbers that satisfy this form, $g(N)=2^R$ . The sum of $g(N)$ for all $N$ in this form and interval is $2^{k-1}$ $R$ can vary between $1$ and $k-1$ , so the total sum is $\underbrace{2^{k-1}+2^{k-1}\cdots 2^{k-1}}_{k-1}=(k-1)2^{k-1}$ . The domain of the function we are trying to find is all even integers in the interval $[2^1, 2^n]$ so there are $n-1$ values of $k$ . Now we have the sum $\sum_{k=1}^{n-1}{(k-1)2^{k-1}}=\sum_{i=1}^{n-2}{k\cdot2^{k}}$ . However, we did not consider powers of two yet(since our interval was strictly between powers of 2), so we have to add $\sum_{k=1}^{n}{2^k}$ .
Note that $\sum_{i=1}^{n-2}{k\cdot2^{k}}=(2^1+2^2\cdots 2^{n-2})+(2^2+2^3\cdots 2^{n-2})\cdots +(2^{n-3}+2^{n-2})+2^{n-2}=2^1(2^{n-2}-1)+2^2(2^{n-3}-1)\cdots +2^{n-3}(2^2-1)+2^{n-2}(2^1-1)=(n-2)(2^{n-1})-\sum_{k=1}^{n-2}{2^k}=(n-2)(2^{n-1})-2(2^{n-2}-1)=(n-3)(2^{n-1})+2$ . Adding $\sum_{k=1}^{n}{2^k}=2(2^n-1)$ , we get $(n+1)(2^{n-1})$ .
If this sum is a perfect square, $n\not\equiv0\pmod2$ , since that would make $2^{n-1}$ not a perfect square, and $n+1$ odd so it cannot contribute a factor of 2 to make the power of 2 a perfect square.
We want the least odd number less than $1000$ such that $(n+1)$ is an even perfect square. The greatest even square less than $1000$ is $30^2=900$ so $n+1=900 \Longrightarrow n=\boxed{899}$
| 899
|
6,315
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_I_Problems/Problem_14
| 1
|
A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$
|
Diagram borrowed from Solution 1
Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields \[BO=\sqrt{TB^2-TO^2}=3\] Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and \[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\] Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields \[TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}\] Apply Law of Cosines on $\bigtriangleup TBC$ we have \[BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC\] \[(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC\] \[\cos BTC=\frac{23}{50}\] Apply Law of Cosines on $\bigtriangleup STB$ using the fact that $\angle STB=\angle BTC$ we have \[SB^2=ST^2+BT^2-2(ST)(BT)\cos STB\] \[SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}\] Apply Pythagorean Theorem on $\bigtriangleup BSM$ yields \[SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}\] Let the perpendicular from $T$ hits $SBC$ at $P$ . Let $SP=x$ and $PM=\frac{\sqrt{1585}}{10}-x$ . Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \[TP^2=TS^2-SP^2=TM^2-PM^2\] \[4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2\] Cancelling out the $x^2$ term and solving gets $x=\frac{181}{2\sqrt{1585}}$
Finally, by Pythagorean Theorem, \[TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}\] so $\lfloor m+\sqrt{n}\rfloor=\boxed{183}$
| 183
|
6,316
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_2
| 1
|
The lengths of the sides of a triangle with positive area are $\log_{10} 12$ $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$
|
By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$ , we have that
Also,
Combining these two inequalities:
\[6.25 < n < 900\]
Thus $n$ is in the set $(6.25 , 900)$ ; the number of positive integer $n$ which satisfies this requirement is $\boxed{893}$
| 893
|
6,317
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_3
| 2
|
Let $P$ be the product of the first $100$ positive odd integers . Find the largest integer $k$ such that $P$ is divisible by $3^k .$
|
We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number $\equiv 2 \pmod{3}$ in set S will be divisible by 3. In other words, every number $\equiv 3 \pmod{6}$ . This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Using some simple math, there are 33 numbers that fit this description. All you have to do is find the largest odd number that is divisible by 3 and below 200, then add 3 and divide by 6. Next, we find the number of odd digits below 200 that are divisible by 9. The same strategy works, and gives us 11. 27 gives 3, and 81 gives 1. $33 + 11 + 4 + 1 = \boxed{49}$
| 49
|
6,318
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5
| 1
|
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$
|
Without loss of generality , assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$ , totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$ . Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$
\[A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}\]
Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so
\begin{align*}A(1)\cdot A(6)+A(1)\cdot A(6)&=\frac{5}{96}\\ A(1)\cdot A(6)&=\frac{5}{192}\end{align*}
Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$ , so:
\[A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6} \Longrightarrow A(1)+A(6)=\frac{1}{3}\]
Combining the equations:
\begin{align*}A(6)\left(\frac{1}{3}-A(6)\right)&=\frac{5}{192}\\ 0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\\ A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*} We know that $A(6)>\frac{1}{6}$ , so it can't be $\frac{1}{8}$ . Therefore, the probability is $\frac{5}{24}$ and the answer is $5+24=\boxed{29}$
| 29
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6,319
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5
| 2
|
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$
|
We have that the cube probabilities to land on its faces are $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}+x$ $\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \[4 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}\] multiplying by 288 we get: \[32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15\] dividing by 16 and rearranging we get: \[\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}\] so the probability F which is greater than $\frac{1}{6}$ is equal $\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{29}$
| 29
|
6,320
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_5
| 3
|
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$
|
Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is \[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\] Let the probability of obtaining face $F$ be $(1/6)+x$ . Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$ . Therefore
\begin{align*} {{47}\over{288}}&= 4\left({1\over6}\right)^2+2\left({1\over6}+x\right) \left({1\over6}-x\right)\cr&= {4\over36}+2\left({1\over36}-x^2\right)\cr&= {1\over6}-2x^2. \end{align*}
Then $2x^2=1/288$ , and so $x=1/24$ . The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$ , and $m+n=\boxed{29}$
| 29
|
6,321
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6
| 1
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Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$
|
[asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp}; draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$E$", Ep, SE); label("$F$", F, E); label("$A'$", Ap, W); label("$C'$", Cp, SW); label("$D'$", Dp, E); label("$s$", Ap--B, W); label("$1$", A--D, N); [/asy] Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$ , and define $s$ to be one of the sides of that square. Since the sides are parallel , by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$ . Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$
$\angle EAF$ is $60$ degrees, so $\angle BAE = \frac{90 - 60}{2} = 15$ . Thus, $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}$ , so $AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}$ . Since $\triangle AEF$ is equilateral $EF = AE = \sqrt{6} - \sqrt{2}$ $\triangle CEF$ is a $45-45-90 \triangle$ , so $CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1$ . Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \sqrt{3} - s)$ , so $(3 - \sqrt{3})s = 2 - \sqrt{3}$ . Therefore, $s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}$ , and $a + b + c = 3 + 3 + 6 = \boxed{12}$
| 12
|
6,322
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6
| 2
|
Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$
|
Suppose $\overline{AB} = \overline{AD} = x.$ Note that $\angle EAF = 60$ since the triangle is equilateral, and by symmetry, $\angle BAE = \angle DAF = 15.$ Note that if $\overline{AD} = x$ and $\angle BAE = 15$ , then $\overline{AA'}=\frac{x}{\tan(15)}.$ Also note that \[AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x\] Using the fact $\tan(15) = 2-\sqrt{3}$ , this yields \[x = \frac{1}{3+\sqrt{3}} = \frac{3-\sqrt{3}}{6} \rightarrow 3 + 3 + 6 = \boxed{12}\]
| 12
|
6,323
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_6
| 3
|
Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$
|
Why not solve in terms of the side $x$ only (single-variable beauty)? By similar triangles we obtain that $BE=\frac{x}{1-x}$ , therefore $CE=\frac{1-2x}{1-x}$ . Then $AE=\sqrt{2}*\frac{1-2x}{1-x}$ . Using Pythagorean Theorem on $\triangle{ABE}$ yields $\frac{x^2}{(1-x)^2} + 1 = 2 * \frac{(1-2x)^2}{(1-x)^2}$ . This means $6x^2-6x+1=0$ , and it's clear we take the smaller root: $x=\frac{3-\sqrt{3}}{6}$ . Answer: $\boxed{12}$
| 12
|
6,324
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7
| 1
|
Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.
|
There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting ).
Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$ . Therefore, there are $999 - (99 + 162) = \boxed{738}$ such ordered pairs.
| 738
|
6,325
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7
| 2
|
Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.
|
We proceed by casework on the number of digits of $a.$
Case 1: Both $a$ and $b$ have three digits
We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \cdot 8 \cdot 9 = 576$ possible values in this case.
Case 2: $a$ (or $b$ ) has two digits
If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$ ). There thus are $8 \cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \cdot 2 = 144$ possible values in this case.
Case 3: $a$ (or $b$ ) has one digit
This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \cdot 2 = 18$ possible values for this case.
Adding them all up, we get $576 + 144 + 18 = \boxed{738},$ and we're done.
| 738
|
6,326
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_7
| 3
|
Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.
|
For every $a \in [1, 999]$ $(a, 1000 - a)$ is a potential candidate for a solution, barring the cases where $a$ or $1000 -a$ has zero digits.
First, let's consider all viable $a$ that do not have a zero digit. As there are 9 non-zero digits, we have:
However, we have still overlooked the cases where $1000 - a$ contains zero digits:
Adding up the cases with $a \in [1, 999]$ with no zero digits and removing the cases with $1000 - a$ with zero digits gives us \[(9^1 + 9^2 + 9^3) - 9^2 = \boxed{738}\]
| 738
|
6,327
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8
| 1
|
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]
|
If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.
There are 6 possible colors for the center triangle.
Thus, in total we have $6\cdot(20 + 30 + 6) = \boxed{336}$ total possibilities.
| 336
|
6,328
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8
| 2
|
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]
|
We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.
Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions.
Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this.
Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this.
Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all.
Now, we add our cases up: $1296+36+36+648=2016$ . We have to divide by 6, so $2016/6=\boxed{336}$ distinguishable ways to color the triangle.
| 336
|
6,329
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_8
| 3
|
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]
|
There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \[6 \binom{5 + 3}{3} = 6\binom{8}{3} = \boxed{336}.\]
| 336
|
6,330
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10
| 1
|
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$
Of these three cases ( $|A| > |B|, |A| < |B|, |A|=|B|$ ), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).
There are ${5\choose k}$ ways to $A$ to have $k$ victories, and ${5\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$
Thus $p = \frac 12 \left(1-\frac{126}{512}\right) = \frac{193}{512}$ . The desired probability is the sum of the cases when $|A| \ge |B|$ , so the answer is $\frac{126}{512} + \frac{193}{512} = \frac{319}{512}$ , and $m+n = \boxed{831}$
| 831
|
6,331
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10
| 2
|
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games.
Summing these 6 cases, we get $\frac{638}{1024}$ , which simplifies to $\frac{319}{512}$ , so our answer is $319 + 512 = \boxed{831}$
| 831
|
6,332
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10
| 3
|
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
We can apply the concept of generating functions here.
The generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$ . Thus, the total generating function for number of games he wins is
The generating function for $A$ is the same except that it is multiplied by $x$ instead of $(1+0x)$ .
Thus, the generating function for $A$ is
$1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}$
The probability that $B$ wins 0 games is $\frac{1}{32}$ . Since the coefficients for all $x^{n}$ where
$n \geq 1$ sums to 32, the probability that $A$ wins more games is $\frac{32}{32}$
Thus, the probability that $A$ has more wins than $B$ is $\frac{1}{32} \times \frac{32}{32} + \frac{5}{32} \times \frac{31}{32} + \frac{10}{32} \times \frac{26}{32} + \frac{10}{32} \times \frac{16}{32} + \frac{5}{32} \times \frac{6}{32} +\frac{1}{32} \times \frac{1}{32} = \frac{638}{1024} = \frac{319}{512}$
Thus, $319 + 512 = \boxed{831}$
| 831
|
6,333
|
https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_10
| 4
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Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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After the first game, there are $10$ games we care about-- those involving $A$ or $B$ . There are $3$ cases of these $10$ games: $A$ wins more than $B$ $B$ wins more than $A$ , or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs $\binom{5}{0}^2+\binom{5}{1}^2+\binom{5}{2}^2+\binom{5}{3}^2+\binom{5}{4}^2+\binom{5}{5}^2 = \binom{10}{5} = 252$ times, by a special case of Vandermonde's Identity . There are therefore $\frac{1024-252}{2} = 386$ possibilities for each of the other two cases.
If $B$ has more wins than $A$ in its $5$ remaining games, then $A$ cannot beat $B$ overall. However, if $A$ has more wins or if $A$ and $B$ are tied, $A$ will beat $B$ overall. Therefore, out of the $1024$ possibilites, $386+252 = 638$ ways where $A$ wins, so the desired probability is $\frac{638}{1024} = \frac{319}{512}$ , and $m+n=\boxed{831}$
| 831
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6,334
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_11
| 1
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sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.
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Define the sum as $s$ . Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:
Thus $s = \frac{a_{28} + a_{30}}{2}$ , and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\boxed{834}$ .−
| 834
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6,335
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12
| 1
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Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$
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Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$ . Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$ . So $\Delta{GBC} \sim \Delta{EAF}$
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$ . Therefore, the answer is $429+433+3=\boxed{865}$
| 865
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6,336
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12
| 2
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Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$
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Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be $O$ and the origin. Then, $A (0,2)$ $B (-\sqrt{3}, -1)$ $C (\sqrt{3}, -1)$ $D$ and $E$ can be calculated easily knowing $AD$ and $AE$ $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$ $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$ . As $DF$ and $EF$ are parallel to $AE$ and $AD$ $F (-1, -12\sqrt{3}+2)$ $G$ and $A$ is the intersection between $AF$ and circle $O$ . Therefore $G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})$ . Using the Shoelace Theorem, $[CBG] = \dfrac{429\sqrt{3}}{433}$ , so the answer is $\boxed{865}$ . Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation.
| 865
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6,337
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12
| 3
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Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$
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Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$ . Since $BAC$ is equilateral with angle of $60^{\circ}$ , angle $D$ is $120^{\circ}$ . Use law of cosines to find $AF = \sqrt{433}$ . Then use law of sines to find angle $BAG$ and $GAC$ . Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$ . Next we use law of cosine on triangles $BAG$ and $GAC$ , solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$ . Next, $AG = 2\cdot R \cos{\theta}$ , where R is radius of circle $= 2$ and $\theta =$ angle $BAG$ . We already know sine of the angle so find cosine, hence we have found $AG$ . At this point it is system of equation yielding $CG = \frac{26\sqrt{3}}{\sqrt{433}}$ and $BG = \frac{22\sqrt{3}}{\sqrt{433}}$ . Given $[CBG] = \frac{BC \cdot CG \cdot BG}{4R}$ , and $BC = 2\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \frac{429\sqrt{3}}{433}$ , to give answer = $\boxed{865}$
| 865
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6,338
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12
| 4
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Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$
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Note that $AB=2\sqrt3$ $DF=11$ , and $EF=13$ . If we take a homothety of the parallelogram with respect to $A$ , such that $F$ maps to $G$ , we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$ . Since $\angle AGB=\angle AGC=60^{\circ}$ , from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$ . Let $BG=11k$ and $CG=13k$
By Law of Cosines on $\triangle BGC$ , we have \[12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}\] Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$
| 865
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6,339
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https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_13
| 1
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How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$
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Let the first odd integer be $2n+1$ $n\geq 0$ . Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$ . The odd integers form an arithmetic sequence with sum $N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$ . Thus, $j$ is a factor of $N$
Since $n\geq 0$ , it follows that $2n+j \geq j$ and $j\leq \sqrt{N}$
Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors of $N$ . This means $N=p_1^2p_2^2$ or $N=p_1p_2^4$ . Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$
Instead we do some casework:
\[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\]
\[N = 4k(n+k) \Longrightarrow \frac{N}{4} = k(n+k)\]
\[\frac{N}{4} = (2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\]
The total number of integers $N$ is $5 + 10 = \boxed{15}$
| 15
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6,340
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_1
| 1
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Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function ).
2005 AIME I Problem 1.png
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Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon . If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles . The length of each side of the triangle is $2r$ . Notice that the radius of circle $C$ is equal to the length of the side of the triangle plus $r$ . Thus, the radius of $C$ has a length of $3r = 30$ , and so $r = 10$ $K = 30^2\pi - 6(10^2\pi) = 300\pi$ , so $\lfloor 300\pi \rfloor = \boxed{942}$
| 942
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6,341
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_3
| 1
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How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?
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Suppose $n$ is such an integer . Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$
In the first case, the three proper divisors of $n$ are $1$ $p$ and $q$ . Thus, we need to pick two prime numbers less than $50$ . There are fifteen of these ( $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$ ) so there are ${15 \choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ numbers of the first type.
In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$ . Thus we need to pick a prime number whose square is less than $50$ . There are four of these ( $2, 3, 5,$ and $7$ ) and so four numbers of the second type.
Thus there are $105+4=\boxed{109}$ integers that meet the given conditions.
| 109
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6,342
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4
| 1
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The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.
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If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ . If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ . Thus, if the number of columns is $n$ , the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$ . In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$ , so this number works and no larger number can. Thus, the answer is $\boxed{294}$
| 294
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6,343
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4
| 2
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The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.
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Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$
To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$
| 294
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6,344
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4
| 3
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The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.
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Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the "extra" pile to be placed at the end. To maximize the number of "extra" people to fill in the last few rows. We remove 3 columns and add 4 rows. For the first new row, one more person will be discarded. For the second, 1 extra person are added since there is one more row now, and there are 2 less columns. Thus, there are 3 extra people discarded. Similarly, 5 extra people are discarded for the third column. Now there are $5+1+3+5=14$ people in the extra pile to put as the last row, so there are $14(14+7)=\boxed{294}$ people.
| 294
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6,345
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4
| 4
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The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.
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Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).
We can see that the number of members in the band must be of the form $n(n + 7)$ for some positive integer $n$ . When $n = 28$ , this product is $980$ , and since AIME answers are nonnegative integers less than $1000$ , we don't have to check any higher $n$ . Also, we know that this product must be 5 more than a perfect square, so we can make a table as shown and bash: \[\begin{tabular}{|c|c|c|} n & n(n+7) & 5 more than a perfect square?\\ \hline 1 & 8 & no\\ \hline 2 & 18 & no\\ \hline 3 & 30 & yes\\ \hline 4 & 44 & no\\ \hline 5 & 60 & no\\ \hline 6 & 78 & no\\ \hline 7 & 98 & no\\ \hline 8 & 120 & no\\ \hline 9 & 144 & no\\ \hline 10 & 170 & no\\ \hline 11 & 198 & no\\ \hline 12 & 228 & no\\ \hline 13 & 260 & no\\ \hline 14 & 294 & yes\\ \hline 15 & 330 & no\\ \hline 16 & 368 & no\\ \hline 17 & 408 & no\\ \hline 18 & 450 & no\\ \hline 19 & 494 & no\\ \hline 20 & 540 & no\\ \hline 21 & 588 & no\\ \hline 22 & 638 & no\\ \hline 23 & 690 & no\\ \hline 24 & 744 & no\\ \hline 25 & 800 & no\\ \hline 26 & 858 & no\\ \hline 27 & 918 & no\\ \hline 28 & 980 & no\\ \hline \end{tabular}\] Thus, we can see that our largest valid $n(n+7)$ is $\boxed{294}$
| 294
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6,346
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5
| 1
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Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.
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There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70 \cdot 9 = \boxed{630}$
| 630
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6,347
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5
| 2
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Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.
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First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). In the end, we multiply these values together.
First, there are obviously $\binom{8}{4}$ ways to order the coins based on color.
Next, we set up a recurrence. Let $a_n$ be the number of ways $n$ indistinguishable coins to be stacked such that no heads are facing each other.
Consider the top coin. If it is facing up, then there are $a_{n-1}$ ways for the rest of the coins below it to be arranged. If it is facing down however, there will only be one way to arrange the coins. We can the recurrence: \[a_n=a_{n-1}+1.\] Note $a_1=2$ , so $a_8 = 9$
Finally, to get the result, we do $70\times 9$ to get $\boxed{630}$
| 630
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6,348
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7
| 1
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In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$
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[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); label("$12$",(10.935,7.794),S); label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E); [/asy]
Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$ . Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$ , and $BC=8$ and $EC=2$ . We are given that $DC=12$ . Since $\angle CED = 120^{\circ}$ , using Law of Cosines on $\bigtriangleup CED$ gives \[12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})\] which gives \[144-4=DE^2+2DE\] . Adding $1$ to both sides gives $141=(DE+1)^2$ , so $DE=\sqrt{141}-1$ $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$ , so $AP=5$ and $BQ=5$ $PQ=DE$ , and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$
| 150
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6,349
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7
| 2
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In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$
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Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles , so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $\triangle DGC$ $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ . The Pythagorean Theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$ , so $EF = GC = \sqrt{141}$ . Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$ , and $p + q = \boxed{150}$
| 150
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6,350
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_8
| 1
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The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots . Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$
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Let $y = 2^{111x}$ . Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ . Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$ . Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ . Then the previous statement says that $2^{111\cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \frac{2}{111}$ . Thus the answer is $111 + 2 = \boxed{113}$
| 113
|
6,351
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11
| 1
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semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$
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We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\sqrt{2}$ right triangle. This means that we have $r + r\sqrt{2} = 8\sqrt{2}$ so $r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}$ . Then the diameter is $32 - \sqrt{512}$ giving us $32 + 512 = \boxed{544}$
| 544
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6,352
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11
| 2
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semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$
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We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives $64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})$ Simplifying gives $d-16\sqrt{2}+d\sqrt{2}=0$ . Solving gives $d=32-16\sqrt{2}=32-\sqrt{512}$ so the answer is $32+512= \boxed{544}$
| 544
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6,353
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11
| 3
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semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$
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It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$ , we can see $x=\sqrt{2}(8-x)$ and we get $2x=32-\sqrt{512}$ and we have the answer $\boxed{544}$ ~bluesoul
| 544
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6,354
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_14
| 1
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Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$
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Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ $ax+by=8a$ $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$ , hence $8a+0b-12b=-4b-7a-10b+9a$ , or $-3a=b$ . We can take $a=-1$ and $b=3$ . So the side of the square is $\frac{44}{\sqrt{10}}$ , the area is $K=\frac{1936}{10}$ , and the answer to the problem is $\boxed{936}$
| 936
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6,355
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2
| 1
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A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$
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Use construction . We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.
Our answer is thus $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}$ , and $m + n = \boxed{79}$
| 79
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6,356
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2
| 2
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A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$
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Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216$ different manners. The total number of possible strings is $\frac{9!}{3!3!3!} = 1680$ . The solution is therefore $\frac{216}{1680} = \frac{9}{70}$ , and $m + n = \boxed{79}$
| 79
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6,357
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2
| 3
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A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$
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The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to ${9 \choose 3}{6 \choose3}$ as the amount of ways to select three rolls out of 9 to give to the first person is ${9 \choose 3}$ , and three rolls out of 6 is ${6 \choose3}$ . After that, the three remaining rolls have no more configurations.
The numerator is the amount of ways to give one roll of each type to each of the three people, which can be done by defining the three types of rolls as x flavored, y flavored, and z flavored.
xxx, yyy, zzz
So you have to choose one x, one y, and one z to give to the first person. There are 3 xs, 3 ys, and 3 zs to select from, giving $3^3$ combinations. Multiply that by the combinations of xs, ys, and zs for the second person, which is evidently $2^3$ since there are two of each letter left.
$(27*8)/{9 \choose 3}{6 \choose3}$ simplifies down to our fraction m/n, which is $9/70$ . Adding them up gives $9 + 70 = \boxed{79}$
| 79
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6,358
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_3
| 1
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An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers . Find $m+n.$
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Let's call the first term of the original geometric series $a$ and the common ratio $r$ , so $2005 = a + ar + ar^2 + \ldots$ . Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$ . Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$ . We know this series has sum $20050 = \frac{a^2}{1 - r^2}$ . Dividing this equation by $\frac{a}{1-r}$ , we get $10 = \frac a{1 + r}$ . Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$ $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$ , so the answer is $399 + 403 = \boxed{802}$
| 802
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6,359
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_4
| 1
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Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$
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$10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors
$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.
$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.
Now, we use the Principle of Inclusion-Exclusion . We have $121 + 64 + 276$ total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers. Thus, we must subtract off the divisors of their pair-wise greatest common divisors
$\gcd(10^{10},15^7) = 5^7$ which has 8 divisors.
$\gcd(15^7, 18^{11}) = 3^7$ which has 8 divisors.
$\gcd(18^{11}, 10^{10}) = 2^{10}$ which has 11 divisors.
So now we have $121 + 64 + 276 - 8 -8 -11$ potential divisors. However, we've now undercounted those factors which divide all three of our numbers. Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of $121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}$
| 435
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6,360
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_4
| 2
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Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$
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We can rewrite the three numbers as $10^{10} = 2^{10}\cdot 5^{10}$ $15^7 = 3^7\cdot5^7$ , and $18^{11} = 2^{11}\cdot3^{22}$ .
Assume that $n$ (a positive integer) is a divisor of one of the numbers. Therefore, $n$ can be expressed as ${p_1}^{e_1}$ or as ${p_2}^{e_2}{p_3}^{e_3}$ where $p_1$ $p_2$ are in $\{2,3,5\}$ and $e_1$ $e_2$ are positive integers.
If $n$ is the power of a single prime, then there are 11 possibilities ( $2^1$ to $2^{11}$ ) for $p_1=2$ , 22 possibilities ( $3^1$ to $3^{22}$ ) for $p_1=3$ , 10 possibilities ( $5^1$ to $5^{10}$ ) for $p_1=5$ , and 1 possibility if $n=1$ . From this case, there are $11+22+10+1=44$ possibilities.
If $n$ is the product of the powers of two primes, then we can just multiply the exponents of each rewritten product to get the number of possibilities, since each exponent of the product must be greater than 0. From this case, there are $10*10+11*22+7*7=100+242+49=391$ possibilities.
Adding up the two cases, there are $44+391=\boxed{435}$ positive integers.
| 435
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6,361
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5
| 2
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Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$
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Let $k=\log_a b$ . Then our equation becomes $k+\frac{6}{k}=5$ . Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$ . Hence $a^2=b$ or $a^3=b$
For the first case $a^2=b$ $a$ can range from 2 to 44, a total of 43 values.
For the second case $a^3=b$ $a$ can range from 2 to 12, a total of 11 values.
Thus the total number of possible values is $43+11=\boxed{54}$
| 54
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6,362
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5
| 3
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Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$
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Using the change of base formula on the second equation to change to base $a$ , we get $\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}$ . If we substitute $x$ for $\log_a(b)$ , we get $x + \frac{6}{x}$ . Multiplying by $x$ on both sides and solving, we get $x=3,2$ . Substituting back in, we get $\log_a(b) = 3,2$ . That means $a^3 = b$ or $a^2 = b$ . Since $b \leq 2005$ , we can see that for the cubed case, the maximum $a$ can be without exceeding 2005 is 12(because $13^3 = 2197$ ) and for the squared case it can be a maximum of 44. Since $a \neq 1$ , the number of values is $(44-1)+(12-1) = \boxed{54}$
| 54
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6,363
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6
| 1
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The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position.
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Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly $131 - 1 = 130$ numbers must be in front of it. There are $\frac{130}{2} = 65$ cards from each of piles A, B in front of card 131. This suggests that $n = 131 + 65 = 196$ ; the total number of cards is $196 \cdot 2 = \boxed{392}$
| 392
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6,364
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6
| 2
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The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position.
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If you index the final stack $1,2,\dots,2n$ , you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers $1,2,\dots, n$ . This all leads us to the following equation. \[131=2n-2(131)+1\implies2n=\boxed{392}\]
| 392
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6,365
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7
| 1
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Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$
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We note that in general,
It now becomes apparent that if we multiply the numerator and denominator of $\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $(\sqrt[16]{5} - 1)$ , the denominator will telescope to $\sqrt[1]{5} - 1 = 4$ , so
It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}$
| 125
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6,366
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7
| 2
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Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$
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Like Solution $2$ , let $z=\sqrt[16]{5}$ Then, the expression becomes
$x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}$ Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
$x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}$ Plugging back in,
$x=\frac{4({\sqrt[16]{5}-1)}}{5-1}\implies x=\sqrt[16]{5}-1$
Hence, after some basic exponent rules, we find the answer is $\boxed{125}$
| 125
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6,367
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8
| 1
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Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime , and $n$ is not divisible by the square of any prime , find $m+n+p.$
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Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$ . Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$ , respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$ . Let the endpoints of the chord/tangent be $A,B$ , and the foot of the perpendicular from $O_3$ to $\overline{AB}$ be $T$ . From the similar right triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$
\[\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.\]
It follows that $HO_1 = \frac{28}{3}$ , and that $O_3T = \frac{58}{7}\dagger$ . By the Pythagorean Theorem on $\triangle ATO_3$ , we find that
\[AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}\]
and the answer is $m+n+p=\boxed{405}$
| 405
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6,368
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8
| 2
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Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime , and $n$ is not divisible by the square of any prime , find $m+n+p.$
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Call our desired length $x$ . Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$ , we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$ . Then, extend $\overline{O_3T}$ until it reaches the circle on both sides; call them $P,Q$ . By Power of a Point, $\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}$ .
Since $\overline{AT}=\overline{TB}=\frac{1}{2}x$ \[(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2\] \[\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2\] After solving for $x$ , we get $x=\frac{8\sqrt{390}}{7}$ , so our answer is $8+390+7=\boxed{405}$
| 405
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6,369
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 1
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For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
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We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$ . So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities $\cos \left(\frac{\pi}2 - u\right) = \sin u$ and $\sin \left(\frac{\pi}2 - u\right) = \cos u$ hold for all real $u$ . If our original equation holds for all $t$ , it must certainly hold for $t = \frac{\pi}2 - u$ . Thus, the question is equivalent to asking for how many positive integers $n \leq 1000$ we have that $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$
$\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$ . We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all $n$ such that $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ hold for all real $u$
$\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$ . So from the equality of the real parts we need either $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$ , in which case $n = 1 + 4k$ , or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$ , in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$ . Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work. There are $\boxed{250}$ of them in the given range.
| 250
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6,370
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 2
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For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
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This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$ . Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$ . Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$ . Hence we must have $i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}$ . Thus since $1000$ is a multiple of $4$ exactly one quarter of the residues are congruent to $1$ hence we have $\boxed{250}$
| 250
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6,371
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 4
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For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
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We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is $cis(n(90^{\circ}-t))$ , and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$ . Canceling gets $n \equiv 1 \mod 4$ , and thus there are $\boxed{250}$ values of n.
| 250
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6,372
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 5
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For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
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Let $t=0$ . Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$ . Thus, the answer is $\boxed{250}$
| 250
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6,373
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 6
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For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
|
We factor out $i^n$ from $(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).$ We know the final expression must be the same as $\sin nt + i \cos nt$ so we must have $i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt$ in which testing yields $n \equiv 1 \pmod{4}$ is the only mod that works, so we have a total of $1000 \cdot\frac{1}{4} = \boxed{250}$ integers $n$ that work.
| 250
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6,374
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 7
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For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
|
Note that this looks like de Moivre's except switched around. Using de Moivre's as motivation we try to convert the given expression into de Moivre's. Note that $\sin t = \cos(90 - t)$ and $\cos t = \sin(90 - t)$ . So we rewrite the expression and setting it equal to the given expression in the problem, we get $\cos(90 - nt) + i\sin(90 - nt) = \cos(90n - nt) + i\sin(90n - nt)$ . Now we can just look at the imaginary parts. Doing so and simplifying, we see that $1 + 4k= n$ . From this we see that $n \equiv 1\pmod{4}$ . So there are $\boxed{250}$ solutions.
| 250
|
6,375
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9
| 8
|
For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$
|
$(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n$
Hence the required condition is just $i^{n} = i$ which is true for exactly 1 in 4 consecutive numbers. Thus $\boxed{250}$
| 250
|
6,376
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11
| 1
|
Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$
|
For $0 < k < m$ , we have
Thus the product $a_{k}a_{k+1}$ is a monovariant : it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$ , so when $k = \frac{37 \cdot 72}{3} = 888$ $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$ , our answer.
| 889
|
6,377
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11
| 2
|
Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$
|
Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$ . Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$ , we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$ . Setting the RHS of this equation equal to $3$ , we find that $m$ must be $\boxed{889}$
| 889
|
6,378
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12
| 1
|
Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$
|
Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $\tan \angle EOG = \frac{x}{450}$ , and $\tan \angle FOG = \frac{y}{450}$
By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$ , we see that \[\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.\] Since $\tan 45 = 1$ , this simplifies to $1 - \frac{xy}{450^2} = \frac{x + y}{450}$ . We know that $x + y = 400$ , so we can substitute this to find that $1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2$
Substituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$ . This is a quadratic with roots $200 \pm 50\sqrt{7}$ . Since $y < x$ , use the smaller root, $200 - 50\sqrt{7}$
Now, $BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}$ . The answer is $250 + 50 + 7 = \boxed{307}$
| 307
|
6,379
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12
| 2
|
Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$
|
[asy] size(3inch); pair A, B, C, D, M, O, X, Y; A = (0,900); B = (900,900); C = (900,0); D = (0,0); M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900); draw(A--B--C--D--cycle); draw(X--O--Y); draw(M--O--A); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",X,N); label("$F$",Y,NNE); label("$O$",O,S); label("$M$",M,N); [/asy] Let the midpoint of $\overline{AB}$ be $M$ and let $FB = x$ , so then $MF = 450 - x$ and $AF = 900 - x$ . Drawing $\overline{AO}$ , we have $\triangle OEF\sim\triangle AOF$ , so \[\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).\] By the Pythagorean Theorem on $\triangle OMF$ \[(OF)^2 = 450^2 + (450 - x)^2.\] Setting these two expressions for $(OF)^2$ equal and solving for $x$ (it is helpful to scale the problem down by a factor of 50 first), we get $x = 250\pm 50\sqrt{7}$ . Since $BF > AE$ , we want the value $x = 250 + 50\sqrt{7}$ , and the answer is $250 + 50 + 7 = \boxed{307}$
| 307
|
6,380
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12
| 3
|
Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$
|
Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with $o = 0, a = -450 + 450i, b = 450 + 450i$ , and $f = x + 450i$ . Since $EF$ = 400, $e = (x-400) + 450i$ . From $\angle{EOF} = 45^{\circ}$ , we can deduce that the rotation of point $F$ 45 degrees counterclockwise, $E$ , and the origin are collinear. In other words, \[\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}\] is a real number. Simplyfying using the fact that $e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}$ , clearing the denominator, and setting the imaginary part equal to $0$ , we eventually get the quadratic \[x^2 - 400x + 22500 = 0\] which has solutions $x = 200 \pm 50\sqrt{7}$ . It is given that $AE < BF$ , so $x = 200 - 50\sqrt{7}$ and \[BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.\]
| 307
|
6,381
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12
| 4
|
Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$
|
[asy] size(250); pair A,B,C,D,O,E,F,G,H,K; A = (0,0); B = (900,0); C = (900,900); D = (0,900); O = (450,450); E = (600,0); F = (150,0); G = (-600,0); H = (450,0); K = (0,270); draw(A--B--C--D--cycle); draw(O--E); draw(O--F); draw(O--G); draw(A--G); draw(O--H); label("O",O,N); label("A",A,S); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,SE); label("F",F,S); label("H",H,SW); label("G",G,SW); label("x",H--E,S); label("K",K,NW); [/asy] Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.
Since $\triangle GOE \sim \triangle OHE$ $\frac{GO}{OE} = \frac{450}{x}$ , and by Angle Bisector Theorem $\frac{GF}{FE} = \frac{450}{x}$ . Thus, $GF = \frac{450 \cdot 400}{x}$ $AF = AH-FH = 50+x$ , and $KA = EB$ (90 degree rotation), and now we can bash on 2 similar triangles $\triangle GAK \sim \triangle GHO$
\[\frac{GA}{AK} = \frac{GH}{OH}\] \[\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}\] I hope you like expanding \[x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}\] \[x^2 - 400x + 22500 = 0\] Quadratic formula gives us \[x = 200 \pm 50 \sqrt{7}\] Since AE < BF \[x = 200 - 50 \sqrt{7}\] Thus, \[BF = 250 + 50 \sqrt{7}\] So, our answer is $\boxed{307}$
| 307
|
6,382
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12
| 5
|
Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$
|
We know that G is on the perpendicular bisector of $EF$ , which means that $EJ=JF=200$ $EG=GF=200\sqrt{2}$ and $GH=250$ . Now, let $HO$ be equal to $x$ . We can set up an equation with the Pythagorean Theorem:
\begin{align*} \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ x^2+62500&=80000 \\ x^2&=17500 \\ x&=50\sqrt{7} \end{align*}
Now, since $IO=450$
\begin{align*} HI&=450-x \\ &=450-50\sqrt{7} \\ \end{align*} \\
Since $HI=AJ$ , we now have:
\begin{align*} BF&=AB-AJ-JF \\ &=900-(450-50\sqrt{7})-200 \\ &=250+50\sqrt{7} \\ \end{align*}
This means that our answer would be $250+50+7=\boxed{307}$
| 307
|
6,383
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_13
| 1
|
Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$
|
We define $Q(x)=P(x)-x+7$ , noting that it has roots at $17$ and $24$ . Hence $P(x)-x+7=A(x-17)(x-24)$ . In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$ . Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$ , where $A$ $(x-17)$ , and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ because the product $(x-17)(x-24)$ will either be too large or not be a divisor of $10$ . We find that $x=19$ and $x=22$ are the only values that allow $(x-17)(x-24)$ to be a factor of $10$ . Hence the answer is $19\cdot 22=\boxed{418}$
| 418
|
6,384
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_13
| 2
|
Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$
|
We know that $P(n)-(n+3)=0$ so $P(n)$ has two distinct solutions so $P(x)$ is at least quadratic. Let us first try this problem out as if $P(x)$ is a quadratic polynomial. Thus $P(n)-(n+3)= an^2+(b-1)n+(c-3)=0$ because $P(n)=an^2+bn+c$ where $a,b,c$ are all integers. Thus $P(x)=ax^2+bx+c$ where $a,b,c$ are all integers. We know that $P(17)$ or $289a+17b+c=10$ and $P(24)$ or $576a+24b+c=17$ . By doing $P(24)-P(17)$ we obtain that $287a+7b=7$ or $41a+b=1$ or $-41a=b-1$ . Thus $P(n)=an^2- (41a)n+(c-3)=0$ . Now we know that $b=-41a+1$ , we have $289a+17(-41a+1)+c=10$ or $408a=7+c$ which makes $408a-10=c-3$ . Thus $P(n)=an^2-(41a)n+(408a-10)=0$ . By Vieta's formulas, we know that the sum of the roots( $n$ ) is equal to 41 and the product of the roots( $n$ ) is equal to $408-\frac{10}{a}$ . Because the roots are integers $\frac{10}{a}$ has to be an integer, so $a=1,2,5,10,-1,-2,-5,-10$ . Thus the product of the roots is equal to one of the following: $398,403,406,407,409,410,413,418$ . Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to $41$ is $\boxed{418}$
| 418
|
6,385
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14
| 1
|
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
|
By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$ , we have
\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} \end{align*}
Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$ . The answer is $q = \boxed{463}$
| 463
|
6,386
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14
| 2
|
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
|
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is $\sqrt{21*6*7*8} = 84$ . We can then use similar triangles with triangle $AQC$ and triangle $DSC$ to find $DS=\frac{24}{5}$ . Consequently, from Pythagorean theorem, $SC = \frac{18}{5}$ and $AS = 14-SC = \frac{52}{5}$ . We can also use the Pythagorean theorem on triangle $AQB$ to determine that $BQ = \frac{33}{5}$
Label $AR$ as $y$ and $RE$ as $x$ $RB$ then equals $13-y$ . Then, we have two similar triangles.
Firstly: $\triangle ARE \sim \triangle ASD$ . From there, we have $\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}$
Next: $\triangle BRE \sim \triangle BQA$ . From there, we have $\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}$
Solve the system to get $x = \frac{2184}{463}$ and $y = \frac{4732}{463}$ . Notice that 463 is prime, so even though we use the Pythagorean theorem on $x$ and $13-y$ , the denominator won't change. The answer we desire is $\boxed{463}$
| 463
|
6,387
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14
| 3
|
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
|
Let $\angle CAD = \angle BAE = \theta$ . Note by Law of Sines on $\triangle BEA$ we have \[\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}\] As a result, our goal is to find $\sin{\angle BEA}$ and $\sin{\theta}$ (we already know $AB$ ).
Let the foot of the altitude from $A$ to $BC$ be $H$ . By law of cosines on $\triangle ABC$ we have \[169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}\] It follows that $AH = \frac{56}{5}$ and $HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}$
Note that by PT on $\triangle AHD$ we have that $AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}$ . By Law of Sines on $\triangle ADC$ (where we square everything to avoid taking the square root) we see \[\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.\] How are we going to find $\sin{\angle BEA}$ though? $\angle BEA$ and $\theta$ are in the same triangle. Applying Law of Sines on $\triangle ABC$ we see that \[\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.\] $\theta$ $\angle B$ , and $\angle BEA$ are all in the same triangle. We know they add up to $180^{\circ}$ . There's a good chance we can exploit this using the identity $\sin{p} = \sin{180^{\circ}-p}$
We have that $\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}$ . Success! We know $\sin{\theta}$ and $\sin{\angle B}$ already. Applying the $\sin$ addition formula we see \[\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.\] This is the last stretch! Applying Law of Sines a final time on $\triangle BEA$ we see \[\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.\] It follows that the answer is $\boxed{463}$
| 463
|
6,388
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14
| 4
|
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
|
Since $AE$ and $AD$ are isogonal with respect to the $A$ angle bisector, we have \[\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.\] To prove this, let $\angle BAE=\angle DAC=x$ and $\angle BAD=\angle CAE=y.$ Then, by the Ratio Lemma, we have \[\frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}\] \[\frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}\] and multiplying these together proves the formula for isogonal lines. Hence, we have \[\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}\] so our desired answer is $\boxed{463}.$
| 463
|
6,389
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14
| 5
|
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
|
Let $ED = x$ , such that $BE = 9-x$ . Since $\overline{AE}$ and $\overline{AD}$ are isogonal, we get $\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)$ , and we can solve to get $x = \frac{1632}{463}$ (and $BE = \frac{2535}{463}$ ). Hence, our answer is $\boxed{463}$ . - Spacesam
| 463
|
6,390
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14
| 6
|
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$
|
Diagram borrowed from Solution 1.
Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have \[AB^2=AC^2+BC^2-2(AC)(BC)\cos C\] Solving gets $\cos C=\frac{3}{5}$ , which implies that \[\sin C=\sqrt{1-\cos C}=\frac{4}{5}\] Applying Stewart's Theorem with cevian $AD$ we have \[(BC)(AD)^2+(BC)(BD)(CD)=(CD)(AB)^2+(BD)(AC)^2\] Solving gets $AD=\frac{4\sqrt{205}}{5}$
Applying Law of Sines on $\bigtriangleup ACD$ to solve for $\sin CAD$ we have \[\frac{AD}{\sin C}=\frac{CD}{\sin CAD}\] Solving gets $\sin CAD=\frac{6\sqrt{205}}{205}$ . Thus $\sin BAE=\sin CAD=\frac{6\sqrt{205}}{205}$
Applying Law of Sines on $\bigtriangleup ABC$ we have \[\frac{AC}{\sin B}=\frac{AB}{\sin C}\] Solving gets $\sin B=\frac{56}{65}$
Applying Stewart's Theorem with cevian $AE$ we have \[(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2\] \[(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2\] Solving gets $AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}$
Finally, applying Law of Sines on $\bigtriangleup BAE$ we have \[\frac{AE}{\sin B}=\frac{BE}{\sin BAE}\] \[\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}\] \[7605BE^2-32342BE+2535=0\] Solving the easy quadratic equation gets $BE=\frac{1632}{463}\Longrightarrow q=\boxed{463}$
| 463
|
6,391
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15
| 1
|
Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$
|
Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$
Let $w_3$ have center $(x,y)$ and radius $r$ . Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$ , and if they are internally tangent, it is $|r_1 - r_2|$ . So we have
\begin{align*} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Solving for $r$ in both equations and setting them equal, then simplifying, yields
\begin{align*} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$
So the locus of points that can be the center of the circle with the desired properties is an ellipse
Since the center lies on the line $y = ax$ , we substitute for $y$ and expand: \[1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.\]
We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$ , so $(-96a)^2 - 4(3+4a^2)(276) = 0$
Solving yields $a^2 = \frac{69}{100}$ , so the answer is $\boxed{169}$
| 169
|
6,392
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15
| 2
|
Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$
|
As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$ . Let $F_1=(-5,12)$ and $F_2=(5,12)$ . If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$ , then $CF_1=16-r$ and $CF_2=4+r$ . Therefore, $CF_1+CF_2=20$ . In particular, the locus of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$
Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$ , and the line $y=ax$ is denoted by $\ell$ . Then we reflect the ellipse over $\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below.
By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$ $T$ , and $F_2'$ are collinear, and similarly, $F_2$ $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$ $F_1F_2=F_1'F_2'=10$ , and $F_1F_2'=F_1'F_2=20$ . Note that $\ell$ bisects $\angle F_1'OF_1$ . We can bisect this angle by bisecting $\angle F_1'OF_2$ and $F_2OF_1$ separately.
We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$ . The height of this from the base of $20$ is $\sqrt{69}$ . Therefore, the complex number $\sqrt{69}+10i$ represents the bisection of $\angle F_1'OF_2$
Similarly, using the 5-12-13 triangles, we easily see that $12+5i$ represents the bisection of the angle $F_2OF_1$ . Therefore, we can add these two angles together by multiplying the complex numbers, finding \[\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.\] Now the point $F_1$ is given by the complex number $-5+12i$ . Therefore, to find a point on line $\ell$ , we simply subtract $\frac{1}{2}\angle F_1'OF_1$ , which is the same as multiplying $-5+12i$ by the conjugate of $(\sqrt{69}+10i)(12+5i)$ . We find \[(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).\] In particular, note that the tangent of the argument of this complex number is $\sqrt{69}/10$ , which must be the slope of the tangent line. Hence $a^2=69/100$ , and the answer is $\boxed{169}$
| 169
|
6,393
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15
| 3
|
Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$
|
We use the same reflection as in Solution 2. As $OF_1'=OF_2=13$ , we know that $\triangle OF_1'F_2$ is isosceles. Hence $\angle F_2F_1'O=\angle F_1'F_2O$ . But by symmetry, we also know that $\angle OF_1T=\angle F_2F_1'O$ . Hence $\angle OF_1T=\angle F_1'F_2O$ . In particular, as $\angle OF_1T=\angle OF_2T$ , this implies that $O, F_1, F_2$ , and $T$ are concyclic.
Let $X$ be the intersection of $F_2F_1'$ with the $x$ -axis. As $F_1F_2$ is parallel to the $x$ -axis, we know that \[\angle TXO=180-\angle F_1F_2T.\tag{1}\] But \[180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}\] By the fact that $OF_1F_2T$ is cyclic, \[\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}\] Therefore, combining (1), (2), and (3), we find that \[\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}\]
By symmetry, we also know that \[\angle F_1TO=\angle OTF_1'.\tag{5}\] Therefore, (4) and (5) show by AA similarity that $\triangle F_1OT\sim \triangle OXT$ . Therefore, $\angle XOT=\angle OF_1T$
Now as $OF_1=OF_2'=13$ , we know that $\triangle OF_1F_2'$ is isosceles, and as $F_1F_2'=20$ , we can drop an altitude to $F_1F_2'$ to easily find that $\tan \angle OF_1T=\sqrt{69}/10$ . Therefore, $\tan\angle XOT$ , which is the desired slope, must also be $\sqrt{69}/10$ . As before, we conclude that the answer is $\boxed{169}$
| 169
|
6,394
|
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15
| 4
|
Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$
|
First, rewrite the equations for the circles as $(x+5)^2+(y-12)^2=16^2$ and $(x-5)^2+(y-12)^2=4^2$ .
Then, choose a point $(a,b)$ that is a distance of $x$ from both circles. Use the distance formula between $(a,b)$ and each of $A$ and $C$ (in the diagram above). The distances, as can be seen in the diagram above are $16-x$ and $4+x$ , respectively. \[(a-5)^2+(b-12)^2=(4+x)^2\] \[(a+5)^2+(b-12)^2=(16-x)^2\] Subtracting the first equation from the second gives \[20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2\] Substituting this into the first equation gives \[a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4\] \[b^2-24b+69+\frac{3a^2}4=0\] Now, instead of converting this to the equation of an eclipse, solve for $b$ and then divide by $a$ \[b=\frac{24\pm\sqrt{300-3a^2}}{2}\] We take the smaller root to minimize $\frac b a$ \[\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}\] Now, let $10\cos\theta=a$ . This way, $\sqrt{100-a^2}=10\sin\theta$ .
Substitute this in. $\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ Then, take the derivative of this and set it to 0 to find the minimum value. $\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}$ Then, use this value of $\sin\theta$ to find the minimum of $\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ to get $\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}$
| 169
|
6,395
|
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_1
| 1
|
The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$
|
A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace$
Now, note that $3\cdot 37 = 111$ so $30 \cdot 37 = 1110$ , and $90 \cdot 37 = 3330$ so $87 \cdot 37 = 3219$ . So the remainders are all congruent to $n - 9 \pmod{37}$ . However, these numbers are negative for our choices of $n$ , so in fact the remainders must equal $n + 28$
Adding these numbers up, we get $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}$
| 217
|
6,396
|
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2
| 1
|
Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$
|
Note that since set $A$ has $m$ consecutive integers that sum to $2m$ , the middle integer (i.e., the median) must be $2$ . Therefore, the largest element in $A$ is $2 + \frac{m-1}{2}$
Further, we see that the median of set $B$ is $0.5$ , which means that the "middle two" integers of set $B$ are $0$ and $1$ . Therefore, the largest element in $B$ is $1 + \frac{2m-2}{2} = m$ $2 + \frac{m-1}{2} > m$ if $m < 3$ , which is clearly not possible, thus $2 + \frac{m-1}{2} < m$
Solving, we get \begin{align*} m - 2 - \frac{m-1}{2} &= 99\\ m-\frac{m}{2}+\frac{1}{2}&=101\\ \frac{m}{2}&=100\frac{1}{2}.\\ m &= \boxed{201}
| 201
|
6,397
|
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2
| 2
|
Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$
|
Let us give the elements of our sets names: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$ . So we are given that \[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\] so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$ (this is because $x = 2 - \frac{m-1}{2}$ so plugging this into $x+(m-1)$ yields $\frac{m+3}{2}$ ). Also, \[m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,\] so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$
Then by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|$ $m$ is a positive integer so we must have $99 = \frac{m - 3}2$ and so $m = \boxed{201}$
| 201
|
6,398
|
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2
| 3
|
Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$
|
The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.
First, we note that for set $A$
\[\frac{m(f + l)}{2} = 2m\]
Where $f$ and $l$ represent the first and last terms of $A$ . This comes from the sum of an arithmetic sequence.
Solving for $f+l$ , we find the sum of the two terms is $4$
Doing the same for set B, and setting up the equation with $b$ and $e$ being the first and last terms of set $B$
\[m(b+e) = m\]
and so $b+e = 1$
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set $A$ has half the number of elements as set $B$ , and the difference between the greatest terms of the two two sequences is $99$ (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where $x$ is the last term of set A:
\[2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)\]
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to $4$ and $1$ respectively (add $x$ and $(-x+4)$ to see what i mean).
Solving this equation we find $x = 102$ . We know the first and last terms have to sum to $4$ so we find the first term of the sequence is $-98$ . Now, the solution is in clear sight, we just find the number of integers between $-98$ and $102$ , inclusive, and it is $m = \boxed{201}$
| 201
|
6,399
|
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2
| 4
|
Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$
|
First, calculate the average of set $A$ and set $B$ . It's obvious that they are $2$ and $1/2$ respectively.
Let's look at both sets. Obviously, there is an odd number of integers in the set with $2$ being in the middle, which means that $m$ is an odd number and that the number of consecutive integers on each side of $2$ are equal. In set $B$ , it is clear that it contains an even number of integers, but since the number in the middle is $1/2$ , we know that the range of the consecutive numbers on both sides will be $(x$ to $0)$ and $(1$ to $-y)$
Nothing seems useful right now, but let's try plugging an odd number, $3$ , for $m$ in set $B$ . We see that there are $6$ consecutive integers and $3$ on both sides of $1/2$ . After plugging this into set $A$ , we find that the set equals \[{1,2,3}\] . From there, we find the absolute value of the difference of both of the greatest values, and get 0.
Let's try plugging in another odd number, $55$ . We see that the resulting set of numbers is $(-54$ to $0)$ , and $(1$ to $55)$ . We then plug this into set $A$ , and find that the set of numbers is $(-25$ to $-29)$ which indeed results in the average being $2$ . We then find the difference of the greatest values to be 26.
From here, we see a pattern that can be proven by more trial and error. When we make $m$ equal to $3$ , then the difference is $0$ whearas when we make it $55$ , then the difference is $26$ $55-3$ equals to $52$ and $26-0$ is just $0$ . We then see that $m$ increases twice as fast as the difference. So when the difference is $99$ , it increased $99$ from when it was $0$ , which means that $m$ increased by $99*2$ which is $198$ . We then add this to our initial $m$ of $3$ , and get $\boxed{201}$ as our answer.
| 201
|
6,400
|
https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2
| 5
|
Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$
|
Let the first term of $A$ be $a$ and the first term of $B$ be $b$ . There are $m$ elements in $A$ so $A$ is $a, a+1, a+2,...,a+m-1$ . Adding these up, we get $\frac{2a+m-1}{2}\cdot m = 2m \implies 2a+m=5$ . Set $B$ contains the numbers $b, b+1, b+2,...,b+2m-1$ . Summing these up, we get $\frac{2b+2m-1}{2}\cdot 2m =m \implies 2b+2m=2$ . The problem gives us that the absolute value of the difference of the largest terms in $A$ and $B$ is $99$ . The largest term in $A$ is $a+m-1$ and the largest term in $B$ is $b+2m-1$ so $|b-a+m|=99$ . From the first two equations we get, we can get that $2(b-a)+m=-3$ . Now, we make a guess and assume that $b-a+m=99$ (if we get a negative value for $m$ , we can try $b-a+m=-99$ ). From here we get that $b-a=-102$ . Solving for $m$ , we get that the answer is $\boxed{201}$
| 201
|
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