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.
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.
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, and . 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 TWO: 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: . Expanding, grouping like terms and factoring, we get: . We know that and are both positive integers, so we need only check values of from to (). Plugging in, the only values of that give integral solutions are and . These gives values of and , respectively. . Hence, the answer is .
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.