Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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== 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> and because <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 <math>\frac{m^2}{99}</math> must be an integer. A quick strategy is to find the smallest such integer <math>m</math> such that <math>\frac{m^2}{99}</math> is an integer. We notice that 99 has a prime factorization of <math>3^2 \cdot 11.</math> Let <math>m^2 = n.</math> Since we need a perfect square and 3 is already squared, we just need to square 11. So <math>3^2 \cdot 11^2</math> gives us 1089 as <math>n</math> and <math>m = \sqrt{1089} = 33.</math> We now get the equation <math>(x-y)(x+y) = 33^2</math>, which we can also write as <math>(x-y)(x+y) = 11^2 \cdot 3^2</math>. A very simple guess assumes that <math>x-y=3^2</math> and <math>x+y=11^2</math> since <math>x</math> and <math>y</math> are positive. Finally, we come to the conclusion that <math>x=65</math> and <math>y=56</math>, so <math>x+y+m</math> <math>=</math> <math>\boxed{\textbf{(E) }154}</math>. | 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> and because <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 <math>\frac{m^2}{99}</math> must be an integer. A quick strategy is to find the smallest such integer <math>m</math> such that <math>\frac{m^2}{99}</math> is an integer. We notice that 99 has a prime factorization of <math>3^2 \cdot 11.</math> Let <math>m^2 = n.</math> Since we need a perfect square and 3 is already squared, we just need to square 11. So <math>3^2 \cdot 11^2</math> gives us 1089 as <math>n</math> and <math>m = \sqrt{1089} = 33.</math> We now get the equation <math>(x-y)(x+y) = 33^2</math>, which we can also write as <math>(x-y)(x+y) = 11^2 \cdot 3^2</math>. A very simple guess assumes that <math>x-y=3^2</math> and <math>x+y=11^2</math> since <math>x</math> and <math>y</math> are positive. Finally, we come to the conclusion that <math>x=65</math> and <math>y=56</math>, so <math>x+y+m</math> <math>=</math> <math>\boxed{\textbf{(E) }154}</math>. | ||
+ | Note that all of the solutions used <math>a+b</math> or <math>a-b</math> as part of their solution. | ||
== See Also == | == See Also == |
Revision as of 10:38, 25 November 2018
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 . 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 .
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 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.
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 |
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