Difference between revisions of "2011 AMC 12B Problems/Problem 21"

Revision as of 22:06, 10 October 2021

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 $1\le a\le 9$ , $0\le b\le 9$ .

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

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

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

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

$|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}$

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 $a$ , $b$ .

Short Cut

We can arrive at $|x-y| = 6\sqrt{11(a^2 - b^2)}$ using the method above. Because we know that $|x-y|$ is an integer, it must be a multiple of 6 and 11. Hence the answer is $66.$

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 $56$ .

Solution 2

Let $(x+y)/2 = 10a + b$ and $\sqrt{xy} = 10b + a$ . By AM-GM we know that $a \ge b$ . Squaring and multiplying by 4 on the first equation we get $x^2 + y^2 + 2xy = 400a^2 + 4b^2 + 80ab$ . Squaring and multiplying the second equation by 4 we get $4xy = 400b^2 + 4a^2 + 80ab$ . Subtracting we get $(x-y)^2 = 396(a^2 - b^2)$ . Note that $396 = 2^2 \cdot 3^2 \cdot 11$ . So to make it a perfect square $a^2 - b^2 = 11$ . By quick difference of squares we see that $a = 6$ and $b = 5$ . So the answer is $3 \cdot 2 \cdot 11 = 66$ . ~coolmath_2018

@@ Line 40: / Line 40: @@
 ==Solution 2==
 Let <math>(x+y)/2 = 10a + b</math> and <math>\sqrt{xy} = 10b + a</math>. By AM-GM we know that <math>a \ge b</math>. Squaring and multiplying by 4 on the first equation we get <math>x^2 + y^2 + 2xy = 400a^2 + 4b^2 + 80ab</math>. Squaring and multiplying the second equation by 4 we get <math>4xy = 400b^2 + 4a^2 + 80ab</math>. Subtracting we get <math>(x-y)^2 = 396(a^2 - b^2)</math>. Note that <math>396 = 2^2 \cdot 3^2 \cdot 11</math>. So to make it a perfect square <math>a^2 - b^2 = 11</math>. By quick difference of squares we see that <math>a = 6</math> and <math>b = 5</math>. So the answer is <math>3 \cdot 2 \cdot 11 = 66</math>.
+~coolmath_2018
 == 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

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