Difference between revisions of "2012 AMC 10A Problems/Problem 22"
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− | == Problem | + | == Problem == |
− | The sum of the first <math>m</math> positive odd integers is 212 more than the sum of the first <math>n</math> positive even integers. What is the sum of all possible values of <math>n</math>? | + | The sum of the first <math>m</math> positive odd integers is <math>212</math> more than the sum of the first <math>n</math> positive even integers. What is the sum of all possible values of <math>n</math>? |
<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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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>. | ||
− | |||
− | + | 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. | |
− | |||
− | Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math> | + | 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>. |
+ | |||
+ | |||
+ | 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, 121\cdot 7,</math> and <math>77\cdot 11</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>, so <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>, hence the answer is <math>\boxed{\textbf{(A)}\ 255}</math>. | ||
+ | |||
+ | ~Edits by BakedPotato66 | ||
==Solution 2== | ==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: | 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: | ||
− | <math>(n+a)^2 = n(n+1) + 212</math> | + | <math>(n+a)^2 = n(n+1) + 212</math>. |
− | <math>n^2 + | + | Expanding, grouping like terms and factoring, we get: |
− | <math> | + | <math>n = \frac{(212 - a^2)}{(2a - 1)}</math>. |
− | <math> | + | |
− | <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>\boxed{\textbf{(A)}\ 255}</math>. |
− | <math>n = (212 | + | |
+ | ==Solution 3== | ||
+ | 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 let us 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\cdot11\cdot11</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 <math>n</math> is <math>211+28+16=\boxed{\textbf{(A)}\ 255}</math>. | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc10a/252 | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | == See Also == | ||
− | + | {{AMC10 box|year=2012|ab=A|num-b=21|num-a=23}} | |
+ | {{MAA Notice}} |
Latest revision as of 00:47, 14 March 2022
Contents
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 , so
can equal
,
, or
, hence the answer is
.
~Edits by BakedPotato66
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.