2005 AMC 10B Problems/Problem 24
Contents
Problem
Let 
and 
be two-digit integers such that is obtained by reversing the digits
of . The integers and satisfy 
for some positive integer 
.
What is 
?
Solution
Let 
, without loss of generality with 
. Then 
. It follows that 
, but 
so 
. Then we have 
. Thus 
is a perfect square. Also, since and 
have the same parity, so is a one-digit odd perfect square, namely 
or 
. The latter case gives 
, which does not work. The former case gives 
, which works, and we have 
.
Solution 2
The first steps are the same as above. Let , where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting 
. This is where the solution diverges.
We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get 
. In order to get a perfect square on the left side, 
must make both prime exponents even. Because the a and b are digits, a simple guess would be that 
(the bigger number) equals 11 while 
is a factor of nine (1 or 9). The correct guesses are 
causing 
and 
. The sum of the numbers is 154 (E).
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
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