Difference between revisions of "2012 AMC 10A Problems/Problem 22"
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Revision as of 23:46, 13 March 2022
Contents
[hide]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 | |
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 | ||
All AMC 10 Problems and Solutions |
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