Difference between revisions of "2011 AMC 12B Problems/Problem 21"
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− | <math>|x-y| = 2\sqrt{99(a^2 - b^2)}</math> | + | <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 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> |
Revision as of 08:54, 5 February 2019
Contents
[hide]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
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 33 because the right side is a multiple of 33. 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
.
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.