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Difference between revisions of "2012 AIME I Problems/Problem 10"
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== Solution ==
 
== Solution ==
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It is apparent that for a perfect square <math>s^2</math> to satisfy the constraints, we must have <math>s^2 - 256 = 1000n</math> or <math>(s+16)(s-16) = 1000n.</math> Now in order for <math>(s+16)(s-16)</math> to be a multiple of <math>1000,</math> at least one of <math>s+16</math> and <math>s-16</math> must be a multiple of <math>5,</math> and since <math>s+16</math> and <math>s-16</math> are in different residue classes mod <math>5,</math> one term must have all the factors and thus must be a multiple of <math>125.</math> Furthermore, each of <math>s+16</math> and <math>s-16</math> must have at least two factors of <math>2,</math> since otherwise <math>(s+16)(s-16)</math> could not possibly be divisible by <math>8.</math> So therefore the conditions are satisfied if either <math>s+16</math> or <math>s-16</math> is divisible by <math>500,</math> or equivalently <math>s = 500n \pm 16.</math> Counting up from <math>n=0</math> to <math>n=5,</math> we see that the tenth value of <math>s</math> is <math>500 \cdot 5 - 16 = 2484</math> and thus <math>t= \frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2012|n=I|num-b=9|num-a=11}}

Revision as of 01:33, 17 March 2012

Problem 10

Let $\mathcal{S}$ be the set of all perfect squares whose rightmost three digits in base $10$ are $256$. Let $\mathcal{T}$ be the set of all numbers of the form $\frac{x-256}{1000}$, where $x$ is in $\mathcal{S}$. In other words, $\mathcal{T}$ is the set of numbers that result when the last three digits of each number in $\mathcal{S}$ are truncated. Find the remainder when the tenth smallest element of $\mathcal{T}$ is divided by $1000$.

Solution

It is apparent that for a perfect square $s^2$ to satisfy the constraints, we must have $s^2 - 256 = 1000n$ or $(s+16)(s-16) = 1000n.$ Now in order for $(s+16)(s-16)$ to be a multiple of $1000,$ at least one of $s+16$ and $s-16$ must be a multiple of $5,$ and since $s+16$ and $s-16$ are in different residue classes mod $5,$ one term must have all the factors and thus must be a multiple of $125.$ Furthermore, each of $s+16$ and $s-16$ must have at least two factors of $2,$ since otherwise $(s+16)(s-16)$ could not possibly be divisible by $8.$ So therefore the conditions are satisfied if either $s+16$ or $s-16$ is divisible by $500,$ or equivalently $s = 500n \pm 16.$ Counting up from $n=0$ to $n=5,$ we see that the tenth value of $s$ is $500 \cdot 5 - 16 = 2484$ and thus $t= \frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}$

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions
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