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 "2011 AMC 12B Problems/Problem 21"

(→‎Solution)
(→‎Solution)
Line 8: Line 8:
 
Answer: (D)
 
Answer: (D)
  
−
\frac{x + y}{2} = 10 a+b<math> for some </math>1\le a\le 9 <math>,</math>0\le b\le 9<math>.
+
<math>\frac{x + y}{2} = 10 a+b</math> for some <math>1\le a\le 9 </math>,<math>0\le b\le 9</math>.
  
−
</math>\sqrt{xy} = 10 b+a<math>
+
<math>\sqrt{xy} = 10 b+a</math>
  
−
</math>100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}<math>
+
<math>100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}</math>
  
−
</math>xy = 100b^2 + 20ab + a^2<math>
+
<math>xy = 100b^2 + 20ab + a^2</math>
  
−
</math>\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)<math>
+
<math>\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)</math>
  
  
 
<br />
 
<br />
−
</math>|x-y| = 2\sqrt{99(a^2 - b^2)}<math>
+
<math>|x-y| = 2\sqrt{99(a^2 - b^2)}</math>
  
−
Note that in order for x-y to be integer, </math>(a^2 - b^2)<math> has to be </math>11n<math> for some perfect square </math>n<math>. Since </math>a<math> is at most </math>9<math>, </math>n = 1<math> or </math>4<math>
+
Note that in order for x-y to be integer, <math>(a^2 - b^2)</math> has to be <math>11n</math> for some perfect square <math>n</math>. Since <math>a</math> is at most <math>9</math>, <math>n = 1</math> or <math>4</math>
  
−
If </math>n = 1<math>, </math>|x-y| = 66<math>, if </math>n = 4<math>, </math>|x-y| = 132<math>. In AMC, we are done. Otherwise, we need to show that </math>a^2 -b^2 = 44<math> is impossible.
+
If <math>n = 1</math>, <math>|x-y| = 66</math>, if <math>n = 4</math>, <math>|x-y| = 132</math>. In AMC, we are done. Otherwise, we need to show that <math>a^2 -b^2 = 44</math> is impossible.
  
−
</math>(a-b)(a+b) = 44<math> -> </math>a-b = 1<math>, or </math>2<math> or </math>4<math> and </math>a+b = 44<math>, </math>22<math>, </math>11<math> respectively. And since </math>a+b \le 18<math>, </math>a+b = 11<math>, </math>a-b = 4<math>, but there is no integer solution for </math>a<math>, </math>b<math>.
+
<math>(a-b)(a+b) = 44</math> -> <math>a-b = 1</math>, or <math>2</math> or <math>4</math> and <math>a+b = 44</math>, <math>22</math>, <math>11</math> respectively. And since <math>a+b \le 18</math>, <math>a+b = 11</math>, <math>a-b = 4</math>, but there is no integer solution for <math>a</math>, <math>b</math>.
  
 
In addition:
 
In addition:
−
Note that </math>11n<math> with </math>n = 1<math> may be obtained with </math>a = 6<math> and </math>b = 5<math> as </math>a^2 - b^2 = 36 - 25 = 11$.
+
Note that <math>11n</math> with <math>n = 1</math> may be obtained with <math>a = 6</math> and <math>b = 5</math> as <math>a^2 - b^2 = 36 - 25 = 11</math>.
  
 
==Sidenote==
 
==Sidenote==

Revision as of 21:11, 20 January 2018

Problem

The arithmetic mean of two distinct positive integers $x$ and $y$ is a two-digit integer. The geometric mean of $x$ and $y$ is obtained by reversing the digits of the arithmetic mean. What is $|x - y|$?

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70$

Solution

Answer: (D)

$\frac{x + y}{2} = 10 a+b$ for some $1\le a\le 9$,$0\le b\le 9$.

$\sqrt{xy} = 10 b+a$

$100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}$

$xy = 100b^2 + 20ab + a^2$

$\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$



$|x-y| = 2\sqrt{99(a^2 - b^2)}$

Note that in order for x-y to be integer, $(a^2 - b^2)$ has to be $11n$ for some perfect square $n$. Since $a$ is at most $9$, $n = 1$ or $4$

If $n = 1$, $|x-y| = 66$, if $n = 4$, $|x-y| = 132$. In AMC, we are done. Otherwise, we need to show that $a^2 -b^2 = 44$ is impossible.

$(a-b)(a+b) = 44$ -> $a-b = 1$, or $2$ or $4$ and $a+b = 44$, $22$, $11$ respectively. And since $a+b \le 18$, $a+b = 11$, $a-b = 4$, but there is no integer solution for $a$, $b$.

In addition: Note that $11n$ with $n = 1$ may be obtained with $a = 6$ and $b = 5$ as $a^2 - b^2 = 36 - 25 = 11$.

Sidenote

It is easy to see that $(a,b)=(6,5)$ is the only solution. This yields $(x,y)=(98,32)$. Their arithmetic mean is $65$ and their geometric mean is $56$.

See also

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 12 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=2011_AMC_12B_Problems/Problem_21&oldid=89904"