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 1
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 value 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 .
Solution 2
The first steps are the same as Solution 1. Let , where we know that a and b are digits (whole numbers less than ).
Like Solution 1, we end up getting . This is where the solution diverges.
We know that the left side of the equation is a perfect square because is an integer. If we factor 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 while is a factor of nine (1 or 9). The correct guesses are causing and . The sum of the numbers is
Solution 3
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.
Solution 4
Continue the same as Solution until we get . Knowing that , we have narrowed down our Pythagorean triples. We know that the other squares should be larger than , so we can start testing.
If we start testing the s, it is fruitless since the closest to would be which is not a Pythagorean triple. We can start by testing out the s, and it turns our that is a Pythagorean triple. Therefore, our answer is = .
~Arcticturn
Video Solution
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=7Y7OX5uVPac ~David
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.