2012 AMC 10A Problems/Problem 22

Problem

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. $(2m + x)(2m - x) = 847 = 7\cdot11^2$ .

Our goal is to find possible values for $x$ , then use the equation above to find $n$ . The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\cdot1, 7\cdot121,$ and $11\cdot77$ . The differences between these factors are $846$ , $114$ , and $66$ - those are all possible values for $2x$ . Thus the possibilities for $x$ 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 $\boxed{\textbf{(A)}\ 255}$ .

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:

$(n+a)^2 = n(n+1) + 212$ . Expanding, grouping like terms and factoring, we get: $n = (212 - a^2)/(2a - 1)$ .

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 $\boxed{\textbf{(A)}\ 255}$ .

Solution 3

Using the closed forms for the sums, we get $m^2=n(n+1)+212$ , or $m^2=n^2+n+212$ . We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$ . Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$ . Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7*11*11$ . The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$ , $7$ and $121$ , $11$ and $77$ . In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$ . The sum of all possible values of $n$ is $211+28+16=255$ .

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2012 AMC 10A Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also