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> | ||
− | ==Solution== | + | ==Solution 1== |
Answer: (D) | Answer: (D) | ||
− | \frac{x + y}{2} = 10 a+b<math> for some < | + | <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> |
− | < | + | 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> |
− | < | + | 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 /> | ||
− | < | + | <math>|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}</math> |
− | Note that | + | 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 < | + | 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>. |
+ | |||
+ | ===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: | ||
− | Note that < | + | 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== | ||
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 | ||
== See also == | == See also == |
Latest revision as of 19:48, 4 November 2023
Problem
The arithmetic mean of two distinct positive integers and
is a two-digit integer. The geometric mean of
and
is obtained by reversing the digits of the arithmetic mean. What is
?
Solution 1
Answer: (D)
for some
,
.
Squaring the first and second equations,
Subtracting the previous two equations,
Note that for x-y to be an integer, has to be
for some perfect square
. Since
is at most
,
or
If ,
, if
,
. In AMC, we are done. Otherwise, we need to show that
, or
is impossible.
->
, or
or
and
,
,
respectively. And since
,
,
, but there is no integer solution for
,
.
Short Cut
We can arrive at using the method above. Because we know that
is an integer, it must be a multiple of 6 and 11. Hence the answer is
In addition:
Note that with
may be obtained with
and
as
.
Sidenote
It is easy to see that is the only solution. This yields
. Their arithmetic mean is
and their geometric mean is
.
Solution 2
Let and
. By AM-GM we know that
. Squaring and multiplying by 4 on the first equation we get
. Squaring and multiplying the second equation by 4 we get
. Subtracting we get
. Note that
. So to make it a perfect square
. From difference of squares, we see that
and
. So the answer is
.
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.