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Difference between revisions of "2015 AIME II Problems/Problem 5"

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==Solution==
 
==Solution==
 
Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions <math>(n-1) \times (n-1)</math>. There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid b is <math>2n(n-1)</math>, and the number of ways to pick two squares out of grid A is <math>\dbinom{n^2}{2}</math>. So, the probability that the two chosen squares are adjacent is <math>\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}</math>. We wish to find the smallest positive integer <math>n</math> such that <math>\frac{4}{n(n+1)} < \frac{1}{2015}</math>, and by inspection the first such <math>n</math> is <math>\boxed{090}</math>.
 
Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions <math>(n-1) \times (n-1)</math>. There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid b is <math>2n(n-1)</math>, and the number of ways to pick two squares out of grid A is <math>\dbinom{n^2}{2}</math>. So, the probability that the two chosen squares are adjacent is <math>\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}</math>. We wish to find the smallest positive integer <math>n</math> such that <math>\frac{4}{n(n+1)} < \frac{1}{2015}</math>, and by inspection the first such <math>n</math> is <math>\boxed{090}</math>.
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==See also==
 +
{{AIME box|year=2015|n=II|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 09:14, 27 March 2015

Problem

Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$.

Solution

Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \times (n-1)$. There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid b is $2n(n-1)$, and the number of ways to pick two squares out of grid A is $\dbinom{n^2}{2}$. So, the probability that the two chosen squares are adjacent is $\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}$. We wish to find the smallest positive integer $n$ such that $\frac{4}{n(n+1)} < \frac{1}{2015}$, and by inspection the first such $n$ is $\boxed{090}$.

See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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