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

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<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70</math>
 
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70</math>
  
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==Solution==
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==Solution 1==
 
Answer: (D)
 
Answer: (D)
  
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<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>\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>
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Squaring the first and second equations, <math>\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2</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>
+
Subtracting the previous two equations, <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 />
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<math>|x-y| = 2\sqrt{99(a^2 - b^2)}</math>
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<math>|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(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 for x-y to be an 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>n=4</math>, or <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>.
 +
 +
===Short Cut===
 +
 +
We can arrive at <math>|x-y| = 6\sqrt{11(a^2 - b^2)}</math> using the method above. Because we know that <math>|x-y|</math> is an integer, it must be a multiple of 6 and 11. Hence the answer is <math>66.</math>
  
 
In addition:
 
In addition:
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==Sidenote==
 
==Sidenote==
 
It is easy to see that <math>(a,b)=(6,5)</math> is the only solution. This yields <math>(x,y)=(98,32)</math>. Their arithmetic mean is <math>65</math> and their geometric mean is <math>56</math>.
 
It is easy to see that <math>(a,b)=(6,5)</math> is the only solution. This yields <math>(x,y)=(98,32)</math>. Their arithmetic mean is <math>65</math> and their geometric mean is <math>56</math>.
 +
 +
==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>. From [[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
 +
 +
 +
==Solution 3 (whee!) ==
 +
Let the 2 digit number be <math>10a+b</math>. We know <math>x+y=20a+2b</math> and <math>\sqrt{xy} = 10b+a</math>.
 +
 +
 +
Thus. <math>x^2+2xy+y^2 = 400a^2+80ab+4b^2</math> and <math>xy = 100b^2+20ab+a^2</math>. Notice that <math>|x-y| = \sqrt{(x-y)^2}</math>. We know that <math>(x-y)^2 = (x+y)^2-4xy = 396a^2-396b^2</math>. Then, <math>|x-y| = 6\sqrt{11}\cdot \sqrt{b^2-a^2}</math>. Clearly, <math>|x-y|</math> must be a multiple of 11, so the only possible answer is <math>\boxed{\textbf{(D) }66}</math>.
 +
 +
-skibbysiggy
  
 
== See also ==
 
== See also ==

Latest revision as of 15:17, 5 September 2024

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 1

Answer: (D)

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

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

Squaring the first and second equations, $\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2$

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

Subtracting the previous two equations, $\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$



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

Note that for x-y to be an 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 $n=4$, or $a^2 -b^2 = 44$ 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 $a^2 - b^2 = 11$. From difference of squares, we see that $a = 6$ and $b = 5$. So the answer is $3 \cdot 2 \cdot 11 = 66$. ~coolmath_2018


Solution 3 (whee!)

Let the 2 digit number be $10a+b$. We know $x+y=20a+2b$ and $\sqrt{xy} = 10b+a$.


Thus. $x^2+2xy+y^2 = 400a^2+80ab+4b^2$ and $xy = 100b^2+20ab+a^2$. Notice that $|x-y| = \sqrt{(x-y)^2}$. We know that $(x-y)^2 = (x+y)^2-4xy = 396a^2-396b^2$. Then, $|x-y| = 6\sqrt{11}\cdot \sqrt{b^2-a^2}$. Clearly, $|x-y|$ must be a multiple of 11, so the only possible answer is $\boxed{\textbf{(D) }66}$.

-skibbysiggy

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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