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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<math>(n+a)^2 = n(n+1) + 212</math>. | <math>(n+a)^2 = n(n+1) + 212</math>. | ||
Expanding, grouping like terms and factoring, we get: | Expanding, grouping like terms and factoring, we get: | ||
− | <math>n = (212 - a^2) | + | <math>n = \frac{(212 - a^2)}{(2a - 1)}</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>. | 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>. | ||
==Solution 3== | ==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 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 | + | 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 == | == See Also == |
Latest revision as of 23:47, 13 March 2022
Contents
[hide]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 |
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