2012 AMC 10A Problems/Problem 22

Problem 22

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

$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$

Solution 1

The sum of the first $m$ odd integers is given by $m^2$ . The sum of the first $n$ even integers is given by $n(n+1)$ .

Thus, $m^2 = n^2 + n + 212$ . Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$ .

Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$ . $n$ is clearly an integer, so $1 - 4(212 - m^2) = 4m^2 - 847$ 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), $4m^2 - 847$ must be odd.

Let $x$ = $\sqrt{4m^2 - 847}$ . (Note that this means that $n = \frac{-1 + x}{2}$ .) This can be rewritten as $x^2 = 4m^2 - 847$ , which can then be rewritten to $4m^2 - x^2 = 847$ . Factor the left side by using the difference of squares. $(2m + x)(2m - x) = 847 = 7*11^2$ .

Our goal is to find possible values for $a$ , then use the equation above to find $n$ . The difference between the factors is $(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.$ We have three pairs of factors, $847*1, 7*121, and 11*77$ . The differences between these factors are $846$ , $114$ , and $66$ - those are all possible values for $2a$ . Thus the possibilities for $a$ are $423$ , $57$ , and $33$ .

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

Solution 2

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

$(n+a)^2 = n(n+1) + 212$ $n^2 + 2an + a^2 = n^2 + n + 212$ $2an + a^2 = n + 212$ $2an - n = 212 - a^2$ $n(2a - 1) = 212 - a^2$ $n = (212 - a^2)/(2a - 1)$ .

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

Page

Toolbox

Search

2012 AMC 10A Problems/Problem 22

Problem 22

Solution 1

Solution 2