# Difference between revisions of "2012 AMC 10A Problems/Problem 24"

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<math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math> | <math> \textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253 </math> | ||

+ | |||

+ | == Solution == | ||

+ | |||

+ | Add the two equations. | ||

+ | |||

+ | <math>2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14</math>. | ||

+ | |||

+ | Now, this can be rearranged: | ||

+ | |||

+ | <math>(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14</math> | ||

+ | |||

+ | and factored: | ||

+ | |||

+ | <math>(a - b)^2 + (a - c)^2 + (b - c)^2 = 14</math> | ||

+ | |||

+ | <math>a</math>, <math>b</math>, and <math>c</math> are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that <math>14 = 9 + 4 + 1</math>. | ||

+ | |||

+ | <math>(a-c)^2 = 9 -> a-c = 3</math>, since <math>a-c</math> is the biggest difference. It is impossible to determine by inspection whether <math>a-b = 2</math> or <math>1</math>, or whether <math>b-c = 1</math> or <math>2</math>. | ||

+ | |||

+ | We want to solve for <math>a</math>, so take the two cases and solve them each for an expression in terms of <math>a</math>. Our two cases are <math>(a, b, c) = (a, a-1, a-3)</math> or <math>(a, a-2, a-3)</math>. Plug these values into one of the original equations to see if we can get an integer for <math>a</math>. | ||

+ | |||

+ | <math>a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011</math>, after some algebra, simplifies to | ||

+ | <math>7a = 2021</math>. 2021 is not divisible by 7, so <math>a</math> is not an integer. | ||

+ | |||

+ | The other case gives <math>a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011</math>, which simplifies to <math>8a = 2024</math>. Thus, <math>a = 253</math> and the answer is <math>\qquad\textbf{(E)}</math>. |

## Revision as of 21:36, 9 February 2012

## Problem 24

Let , , and be positive integers with such that and .

What is ?

## Solution

Add the two equations.

.

Now, this can be rearranged:

and factored:

, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that .

, since is the biggest difference. It is impossible to determine by inspection whether or , or whether or .

We want to solve for , so take the two cases and solve them each for an expression in terms of . Our two cases are or . Plug these values into one of the original equations to see if we can get an integer for .

, after some algebra, simplifies to . 2021 is not divisible by 7, so is not an integer.

The other case gives , which simplifies to . Thus, and the answer is .