2005 AMC 10B Problems/Problem 24
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 ?
Let . The given conditions imply , which implies , and they also imply that both and are nonzero. Then . Since this must be a perfect square, all the exponents in its prime factorization must be even. factorizes into , so . However, the maximum value of is , so . The maximum of is , so . Then we have , so is a perfect square, but the only perfect squares that are within our bound on are and . We know , and, for , adding equations to eliminate gives us . Testing gives us , which is impossible, as and must be digits. Therefore, , and .
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
Once again, the solution is quite similar as the above solutions. Since and are two digit integers, we can write and because , substituting and factoring, we get . Therefore, and must be an integer. A quick strategy is to find the smallest such integer such that is an integer. We notice that 99 has a prime factorization of Let Since we need a perfect square and 3 is already squared, we just need to square 11. So gives us 1089 as and We now get the equation , which we can also write as . A very simple guess assumes that and since and are positive. Finally, we come to the conclusion that and , so . Note that all of the solutions used or as part of their solution.
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