Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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<math>\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154 </math> | <math>\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154 </math> | ||
− | == Solution == | + | == Solution 1 == |
Let <math>x = 10a+b, y = 10b+a</math>. The given conditions imply <math>x>y</math>, which implies <math>a>b</math>, and they also imply that both <math>a</math> and <math>b</math> are nonzero. Then <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. Since this must be a perfect square, all the exponents in its prime factorization must be even. <math>99</math> factorizes into <math>3^2 \cdot 11</math>, so <math>11|(a-b)(a+b)</math>. However, the maximum value of <math>a-b</math> is <math>9-1=8</math>, so <math>11|a+b</math>. The maximum of <math>a+b</math> is <math>9+8=17</math>, so <math>a+b=11</math>. Then we have <math>33^2(a-b) = m^2</math>, so <math>a-b</math> is a perfect square, but the only perfect squares that are within our bound on <math>a-b</math> are <math>1</math> and <math>4</math>. We know <math>a+b=11</math>, and, for <math>a-b=1</math>, adding equations to eliminate <math>b</math> gives us <math>2a=12 \Longrightarrow a=6, b=5</math>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</math>, which is impossible, as <math>a</math> and <math>b</math> must be digits. Therefore, <math>(a,b) = (6,5)</math>, and <math>x+y+m=65+56+33=154\ \mathbf{(E)}</math>. | Let <math>x = 10a+b, y = 10b+a</math>. The given conditions imply <math>x>y</math>, which implies <math>a>b</math>, and they also imply that both <math>a</math> and <math>b</math> are nonzero. Then <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. Since this must be a perfect square, all the exponents in its prime factorization must be even. <math>99</math> factorizes into <math>3^2 \cdot 11</math>, so <math>11|(a-b)(a+b)</math>. However, the maximum value of <math>a-b</math> is <math>9-1=8</math>, so <math>11|a+b</math>. The maximum of <math>a+b</math> is <math>9+8=17</math>, so <math>a+b=11</math>. Then we have <math>33^2(a-b) = m^2</math>, so <math>a-b</math> is a perfect square, but the only perfect squares that are within our bound on <math>a-b</math> are <math>1</math> and <math>4</math>. We know <math>a+b=11</math>, and, for <math>a-b=1</math>, adding equations to eliminate <math>b</math> gives us <math>2a=12 \Longrightarrow a=6, b=5</math>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</math>, which is impossible, as <math>a</math> and <math>b</math> must be digits. Therefore, <math>(a,b) = (6,5)</math>, and <math>x+y+m=65+56+33=154\ \mathbf{(E)}</math>. | ||
Revision as of 19:19, 20 January 2019
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 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 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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