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) | ||
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<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== | + | ===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> | 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> |
Revision as of 05:04, 25 December 2022
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
,
.
Note that in order for x-y to be 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
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
. By quick 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.