A More Difficult AMC Problem

In my January 31 post, I gave you an example of an American Mathematics Competition question. However, as I said in that post, that problem was just medium level. The AMC problem that I am about to show you now is a much harder problem. Its difficulty level would be an 8 out of 10, maybe.

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

Given an unlimited amount of time, anybody could solve this problem, whether through the hard way or easy way. However, one must remember that the AMC is very time consuming, and thus one must solve this in a quick way. In order for that to happen, one must have a few number theory concepts in mind, just simple ones. Or one must find that pattern by himself.

So solve it. After you do that, here’s the two ways you could have answered this question.

SOLUTION ONE: 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). These are two concepts you needed to solve the question easily.

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) = 0Use 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*11^2Our 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*1, 7*121, and 11*77. The differences between these factors are 846114, and 66 – those are all possible values for 2a. Thus the possibilities for x are 42357, and 33Now plug in these values into the equation n = \frac{-1 + x}{2}n can equal 21128, or 16. Add 211 + 28 + 16 = 255. The answer is \boxed{\textbf{(A)}\ 255}.

SOLUTION TWO:  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}.

I myself wasn’t able to solve this, because I didn’t know anything about the sum of the first m odd integers being represented by m^2. Or the one about the first n even integers equaling n(n+1). This question showed me that the AMC is not just about smartness, but also a vast amount of knowledge or at least a genius skill of pattern recognition.


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