2011 AMC 12B Problems/Problem 21
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.