2012 AMC 10A Problems/Problem 22
Contents
Problem
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 ?
Solution 1
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
Solution 2
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 .
Solution 3
a
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc10a/252
~dolphin7
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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