# Difference between revisions of "2012 AMC 10A Problems/Problem 22"

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<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math> | <math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math> | ||

− | == Solution == | + | == Solution 1== |

The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>. | The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>. | ||

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Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\qquad\textbf{(B)}</math>. | Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\qquad\textbf{(B)}</math>. | ||

+ | |||

+ | ==Solution 2== | ||

+ | |||

+ | As above, start off by noting that the sum of the first <math>m</math> odd integers <math>= m^2</math> and the sum of the first <math>n</math> even integers <math>= n(n+1)</math>. Clearly <math>m > n</math>, so let <math>m = n + a</math>, where <math>a</math> is some positive integer. We have: | ||

+ | |||

+ | <math>(n+a)^2 = n(n+1) + 212</math> | ||

+ | <math>n^2 + 2an + a^2 = n^2 + n + 212</math> | ||

+ | <math>2an + a^2 = n + 212</math> | ||

+ | <math>2an - n = 212 - a^2</math> | ||

+ | <math>n(2a - 1) = 212 - a^2</math> | ||

+ | <math>n = (212 - a^2)/(2a - 1)</math>. | ||

+ | |||

+ | We know that <math>n</math> and <math>a</math> are both positive integers, so we need only check values of <math>a</math> from <math>1</math> to <math>14</math> (<math>14^2 = 196 < 212 < 15^2 = 225). Plugging in, the only values of </math>a<math> that give integral solutions are </math>1, 4,<math> and </math>6<math>. These gives </math>n<math> values of </math>211, 28,<math> and </math>16<math>, respectively. </math>211 + 28 + 16 = 255<math>. Hence, the answer is </math>\qquad\textbf{(B)}$. |

## Revision as of 12:05, 11 February 2012

## Problem 22

The sum of the first positive odd integers is 212 more than the sum of the first positive even integers. What is the sum of all possible values of ?

## Solution 1

The sum of the first odd integers is given by . The sum of the first even integers is given by .

Thus, . Since we want to solve for n, rearrange as a quadratic equation: .

Use the quadratic formula: . is clearly an integer, so must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2), must be odd.

Let = . (Note that this means that .) This can be rewritten as , which can then be rewritten to . Factor the left side by using the difference of squares. .

Our goal is to find possible values for , then use the equation above to find . The difference between the factors is We have three pairs of factors, . The differences between these factors are , , and - those are all possible values for . Thus the possibilities for are , , and .

Now plug in these values into the equation . can equal , , or . Add . The answer is .

## Solution 2

As above, start off by noting that the sum of the first odd integers and the sum of the first even integers . Clearly , so let , where is some positive integer. We have:

.

We know that and are both positive integers, so we need only check values of from to (a1, 4,6n211, 28,16211 + 28 + 16 = 255\qquad\textbf{(B)}$.