Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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The first steps are the same as above. Let <math>x = 10a+b, y = 10b+a</math>, where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting <math>(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. This is where the solution diverges. | The first steps are the same as above. Let <math>x = 10a+b, y = 10b+a</math>, where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting <math>(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. 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 <math>3^2 * 11</math>. In order to get a perfect square on the left side, <math>(a-b)(a+b)</math> must make both prime exponents even. Because the a and b are digits, a simple guess would be that <math>(a+b)</math> (the bigger number) equals 11 while <math>(a-b)</math> is a factor of nine (1 or 9). The correct guesses are <math>a = 6, b = 5</math> causing <math>x = 65, y = 56,</math> and <math>m = 33</math>. The sum of the numbers is | + | 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 <math>3^2 * 11</math>. In order to get a perfect square on the left side, <math>(a-b)(a+b)</math> must make both prime exponents even. Because the a and b are digits, a simple guess would be that <math>(a+b)</math> (the bigger number) equals 11 while <math>(a-b)</math> is a factor of nine (1 or 9). The correct guesses are <math>a = 6, b = 5</math> causing <math>x = 65, y = 56,</math> and <math>m = 33</math>. The sum of the numbers is <math>\boxed{\textbf{(E) }154}</math> |
== See Also == | == See Also == |
Revision as of 19:32, 23 August 2012
Contents
[hide]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
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 |