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

(Problem 22)
(Solution)
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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>
  
== Solution ==
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== Solution 1==
  
 
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>.
 
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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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>.
 
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>.
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==Solution 2==
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As above, start off by noting that the sum of the first <math>m</math> odd integers <math>= m^2</math> and the sum of the first <math>n</math> even integers <math>= n(n+1)</math>. Clearly <math>m > n</math>, so let <math>m = n + a</math>, where <math>a</math> is some positive integer. We have:
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<math>(n+a)^2 = n(n+1) + 212</math>
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<math>n^2 + 2an + a^2 = n^2 + n + 212</math>
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<math>2an + a^2 = n + 212</math>
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<math>2an - n = 212 - a^2</math>
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<math>n(2a - 1) = 212 - a^2</math>
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<math>n = (212 - a^2)/(2a - 1)</math>.
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We know that <math>n</math> and <math>a</math> are both positive integers, so we need only check values of <math>a</math> from <math>1</math> to <math>14</math> (<math>14^2 = 196 < 212 < 15^2 = 225). Plugging in, the only values of </math>a<math> that give integral solutions are </math>1, 4,<math> and </math>6<math>. These gives </math>n<math> values of </math>211, 28,<math> and </math>16<math>, respectively. </math>211 + 28 + 16 = 255<math>. Hence, the answer is </math>\qquad\textbf{(B)}$.

Revision as of 12:05, 11 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 1

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 $\qquad\textbf{(B)}$.

Solution 2

As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$. Clearly $m > n$, so let $m = n + a$, where $a$ is some positive integer. We have:

$(n+a)^2 = n(n+1) + 212$ $n^2 + 2an + a^2 = n^2 + n + 212$ $2an + a^2 = n + 212$ $2an - n = 212 - a^2$ $n(2a - 1) = 212 - a^2$ $n = (212 - a^2)/(2a - 1)$.

We know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ ($14^2 = 196 < 212 < 15^2 = 225). Plugging in, the only values of$a$that give integral solutions are$1, 4,$and$6$. These gives$n$values of$211, 28,$and$16$, respectively.$211 + 28 + 16 = 255$. Hence, the answer is$\qquad\textbf{(B)}$.

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