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

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==Solution 3==
 
==Solution 3==
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Using the closed forms for the sums, we get <math>m^2=n(n+1)+212</math>, or <math>m^2=n^2+n+212</math>. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to complete the square. Our equation is now <math>4m^2=4n^2+4n+848</math>. Complete the square on the right hand side: <math>4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847</math>. Move over the <math>(2n+1)^2</math> and factor to get <math>(2m-2n-1)(2m+2n+1)=847=7*11*11</math>. The second factor is clearly greater than the first, and the only possible factor pairs are 1 and 847, 7 and 121, 11 and 77. In each of these cases, solve for <math>m</math> and <math>n</math> and we find the solutions <math>(m,n)=(212,211), (32,28), (22,16)</math>. The sum of all possible values of n is <math>211+28+16=255</math>.
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Using the closed forms for the sums, we get <math>m^2=n(n+1)+212</math>, or <math>m^2=n^2+n+212</math>. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to complete the square. Our equation is now <math>4m^2=4n^2+4n+848</math>. Complete the square on the right hand side: <math>4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847</math>. Move over the <math>(2n+1)^2</math> and factor to get <math>(2m-2n-1)(2m+2n+1)=847=7*11*11</math>. The second factor is clearly greater than the first, and the only possible factor pairs are <math>1</math> and <math>847</math>, <math>7</math> and <math>121</math>, <math>11</math> and <math>77</math>. In each of these cases, solve for <math>m</math> and <math>n</math> and we find the solutions <math>(m,n)=(212,211), (32,28), (22,16)</math>. The sum of all possible values of n is <math>211+28+16=255</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 17:20, 27 January 2014

Problem

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. $(2m + x)(2m - x) = 847 = 7\cdot11^2$.

Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\cdot1, 7\cdot121,$ and $11\cdot77$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2a$. Thus the possibilities for $x$ 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 $\boxed{\textbf{(A)}\ 255}$.

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 $\boxed{\textbf{(A)}\ 255}$.

Solution 3

Using the closed forms for the sums, we get $m^2=n(n+1)+212$, or $m^2=n^2+n+212$. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to complete the square. Our equation is now $4m^2=4n^2+4n+848$. Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$. Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7*11*11$. The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$, $7$ and $121$, $11$ and $77$. In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$. The sum of all possible values of n is $211+28+16=255$.

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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