Difference between revisions of "2011 AMC 12B Problems/Problem 21"
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==Solution 2== | ==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>. | 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 == | == See also == |
Revision as of 22:06, 10 October 2021
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 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.