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Difference between revisions of "2012 AIME I Problems/Problem 10"

(Created page with "Let <math>\mathcal{S}</math> be the set of all perfect squares whose rightmost three digits in base <math>10</math> are <math>256</math>. Let <math>\mathcal{T}</math> be the set ...")
 
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==Problem 10==
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Let <math>\mathcal{S}</math> be the set of all perfect squares whose rightmost three digits in base <math>10</math> are <math>256</math>. Let <math>\mathcal{T}</math> be the set of all numbers of the form <math>\frac{x-256}{1000}</math>, where <math>x</math> is in <math>\mathcal{S}</math>. In other words, <math>\mathcal{T}</math> is the set of numbers that result when the last three digits of each number in <math>\mathcal{S}</math> are truncated. Find the remainder when the tenth smallest element of <math>\mathcal{T}</math> is divided by <math>1000</math>.
 
Let <math>\mathcal{S}</math> be the set of all perfect squares whose rightmost three digits in base <math>10</math> are <math>256</math>. Let <math>\mathcal{T}</math> be the set of all numbers of the form <math>\frac{x-256}{1000}</math>, where <math>x</math> is in <math>\mathcal{S}</math>. In other words, <math>\mathcal{T}</math> is the set of numbers that result when the last three digits of each number in <math>\mathcal{S}</math> are truncated. Find the remainder when the tenth smallest element of <math>\mathcal{T}</math> is divided by <math>1000</math>.
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== Solution ==
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== See also ==
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{{AIME box|year=2012|n=I|num-b=9|num-a=11}}

Revision as of 00:16, 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

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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