Difference between revisions of "2012 AMC 10A Problems/Problem 22"
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Thus, <math>m^2 = n^2 + n + 212</math>. Since we want to solve for n, rearrange as a quadratic equation: <math>n^2 + n + (212 - m^2) = 0</math>. | Thus, <math>m^2 = n^2 + n + 212</math>. Since we want to solve for n, rearrange as a quadratic equation: <math>n^2 + n + (212 - m^2) = 0</math>. | ||
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− | Let <math>x = \sqrt{4m^2 - 847}</math> | + | Use the quadratic formula: <math>n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}</math>. Since <math>n</math> is clearly an integer, <math>1 - 4(212 - m^2) = 4m^2 - 847</math> must be not only a perfect square, but also an odd perfect square for <math>n</math> to be an integer. |
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+ | Let <math>x = \sqrt{4m^2 - 847}</math>; note that this means <math>n = \frac{-1 + x}{2}</math>. It can be rewritten as <math>x^2 = 4m^2 - 847</math>, so <math>4m^2 - x^2 = 847</math>. Factoring the left side by using the difference of squares, we get <math>(2m + x)(2m - x) = 847 = 7\cdot11^2</math>. | ||
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Our goal is to find possible values for <math>x</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.</math> We have three pairs of factors, <math>847\cdot1, 7\cdot121,</math> and <math>11\cdot77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2x</math>. Thus the possibilities for <math>x</math> are <math>423</math>, <math>57</math>, and <math>33</math>. | Our goal is to find possible values for <math>x</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.</math> We have three pairs of factors, <math>847\cdot1, 7\cdot121,</math> and <math>11\cdot77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2x</math>. Thus the possibilities for <math>x</math> are <math>423</math>, <math>57</math>, and <math>33</math>. | ||
− | Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\boxed{\textbf{(A)}\ 255}</math>. | + | Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. |
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+ | <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. | ||
+ | The answer is <math>\boxed{\textbf{(A)}\ 255}</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 09:31, 26 July 2021
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 .
can equal , , or . Add . The answer is .
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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.