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, because and have the same parity, 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
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AMC 10 Problems and Solutions |
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