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

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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</math>). 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)}</math>.
 
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</math>). 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)}</math>.
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== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=21|num-a=23}}

Revision as of 15:37, 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:

$(n+a)^2 = n(n+1) + 212$. Expanding, grouping like terms and factoring, we get: $n = (212 - a^2)/(2a - 1)$.

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

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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