Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 3)
 
(9 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== 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>
Line 11: Line 11:
 
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>. <math>n</math> is clearly an integer, so <math>1 - 4(212 - m^2) = 4m^2 - 847</math> 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), <math>4m^2 - 847</math> must be odd.
 
  
−
Let <math>x</math> = <math>\sqrt{4m^2 - 847}</math>. (Note that this means that <math>n = \frac{-1 + x}{2}</math>.) This can be rewritten as <math>x^2 = 4m^2 - 847</math>, which can then be rewritten to <math>4m^2 - x^2 = 847</math>. Factor the left side by using the difference of squares. <math>(2m + x)(2m - x) = 847 = 7\cdot11^2</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.
  
−
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, 7\cdot121,</math> and <math>11\cdot77</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>. <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>\boxed{\textbf{(A)}\ 255}</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==
Line 31: Line 36:
 
==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 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>.
 
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

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: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$. Since $n$ is clearly an integer, $1 - 4(212 - m^2) = 4m^2 - 847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.


Let $x = \sqrt{4m^2 - 847}$; note that this means $n = \frac{-1 + x}{2}$. It can be rewritten as $x^2 = 4m^2 - 847$, so $4m^2 - x^2 = 847$. Factoring the left side by using the difference of squares, we get $(2m + x)(2m - x) = 847 = 7\cdot11^2$.


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, $847\cdot1, 121\cdot 7,$ and $77\cdot 11$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2x$. Thus the possibilities for $x$ are $423$, $57$, and $33$.

Now plug in these values into the equation $n = \frac{-1 + x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $\boxed{\textbf{(A)}\ 255}$.

~Edits by BakedPotato66

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: $n = \frac{(212 - a^2)}{(2a - 1)}$.

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 let us 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 $(2m-2n-1)(2m+2n+1)=847=7\cdot11\cdot11$. 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 $211+28+16=\boxed{\textbf{(A)}\ 255}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_22&oldid=172611"