Difference between revisions of "2005 AMC 10B Problems/Problem 24"
Aops12142015 (talk | contribs) (→Solution 3) |
Aops12142015 (talk | contribs) (→Solution 3) |
||
Line 15: | Line 15: | ||
== Solution 3 == | == Solution 3 == | ||
− | Once again, the solution is quite similar as the above solutions. Since <math>x</math> and <math>y</math> are two digit integers, we can write <math>x = 10a+b, y = 10b+a</math>. Since <math>x^2 - y^2 = (x-y)(x+y)</math>, substituting and factoring, we get <math>99(a+b)(a-b) = m^2</math>. Therefore, <math>(a+b)(a-b) = \frac{m^2}{99}</math> and \frac{m^2}{99} must be an integer. Therefore a quick strategy is to find the smallest such integer <math>m</math> such that \frac{m^2}{99} is an integer. We notice that 99 has a prime factorization like <math>3^2 | + | Once again, the solution is quite similar as the above solutions. Since <math>x</math> and <math>y</math> are two digit integers, we can write <math>x = 10a+b, y = 10b+a</math>. Since <math>x^2 - y^2 = (x-y)(x+y)</math>, substituting and factoring, we get <math>99(a+b)(a-b) = m^2</math>. Therefore, <math>(a+b)(a-b) = \frac{m^2}{99}</math> and \frac{m^2}{99} must be an integer. Therefore a quick strategy is to find the smallest such integer <math>m</math> such that \frac{m^2}{99} is an integer. We notice that 99 has a prime factorization like <math>3^2 \cdot 11</math> |
== See Also == | == See Also == |
Revision as of 17:10, 9 June 2016
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
Solution 3
Once again, the solution is quite similar as the above solutions. Since and are two digit integers, we can write . Since , substituting and factoring, we get . Therefore, and \frac{m^2}{99} must be an integer. Therefore a quick strategy is to find the smallest such integer such that \frac{m^2}{99} is an integer. We notice that 99 has a prime factorization like
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.