2012 AMC 10A Problems/Problem 22
The sum of the first positive odd integers is more than the sum of the first positive even integers. What is the sum of all possible values of ?
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: . Since is clearly an integer, must be not only a perfect square, but also an odd perfect square for to be an integer.
Let ; note that this means . It can be rewritten as , so . Factoring the left side by using the difference of squares, we get .
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 , so can equal , , or , hence the answer is .
~Edits by BakedPotato66
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 .
Using the closed forms for the sums, we get , or . 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 . Complete the square on the right hand side: . Move over the and factor to get . The second factor is clearly greater than the first, and the only possible factor pairs are and , and , and . In each of these cases, solve for and and we find the solutions . The sum of all possible values of is .
Video Solution by Richard Rusczyk
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