Difference between revisions of "2012 AMC 10A Problems/Problem 22"

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<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math>
 
<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math>
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== Solution ==
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The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>.
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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>.
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Use the quadratic formula: <math>n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}</math>. <math>n</math> is clearly an integer, so <math>1 - 4(212 - m^2) = 4m^2 - 847</math> 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), <math>4m^2 - 847</math> must be odd.
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Let <math>x</math> = <math>\sqrt{4m^2 - 847}</math>. (Note that this means that <math>n = \frac{-1 + x}{2}</math>.) This can be rewritten as <math>x^2 = 4m^2 - 847</math>, which can then be rewritten to <math>4m^2 - x^2 = 847</math>. Factor the left side by using the difference of squares. <math>(2m + x)(2m - x) = 847 = 7*11^2</math>.
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Our goal is to find possible values for <math>a</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.</math> We have three pairs of factors, <math>847*1, 7*121, and 11*77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2a</math>. Thus the possibilities for <math>a</math> are <math>423</math>, <math>57</math>, and <math>33</math>.
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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>\qquad\textbf{(B)}</math>.

Revision as of 20:21, 9 February 2012

Problem 22

The sum of the first $m$ positive odd integers is 212 more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$?

$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$

Solution

The sum of the first $m$ odd integers is given by $m^2$. The sum of the first $n$ even integers is given by $n(n+1)$.

Thus, $m^2 = n^2 + n + 212$. Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$.

Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$. $n$ is clearly an integer, so $1 - 4(212 - m^2) = 4m^2 - 847$ 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), $4m^2 - 847$ must be odd.

Let $x$ = $\sqrt{4m^2 - 847}$. (Note that this means that $n = \frac{-1 + x}{2}$.) This can be rewritten as $x^2 = 4m^2 - 847$, which can then be rewritten to $4m^2 - x^2 = 847$. Factor the left side by using the difference of squares. $(2m + x)(2m - x) = 847 = 7*11^2$.

Our goal is to find possible values for $a$, then use the equation above to find $n$. The difference between the factors is $(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.$ We have three pairs of factors, $847*1, 7*121, and 11*77$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2a$. Thus the possibilities for $a$ are $423$, $57$, and $33$.

Now plug in these values into the equation $n = \frac{-1 + x}{2}$. $n$ can equal $211$, $28$, or $16$. Add $211 + 28 + 16 = 255$. The answer is $\qquad\textbf{(B)}$.

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