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Difference between revisions of "2005 AMC 10B Problems/Problem 24"

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== Solution 3 ==
 
== Solution 3 ==
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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 <math>\frac{m^2}{99}</math> must be an integer. Therefore 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>
+
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 <math>\frac{m^2}{99}</math> must be an integer. Therefore 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>. So <math>m = \sqrt{1089} = 33.</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 17:15, 9 June 2016

Problem

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$. The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$. What is $x + y + m$?

$\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154$

Solution

Let $x = 10a+b, y = 10b+a$, without loss of generality with $a>b$. Then $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$. It follows that $11|(a-b)(a+b)$, but $a-b < 10$ so $11|a+b \Longrightarrow a+b=11$. Then we have $33^2(a-b) = m^2$. Thus $a-b$ is a perfect square. Also, because $a-b$ and $a+b$ have the same parity, $a-b$ is a one-digit odd perfect square, namely $1$ or $9$. The latter case gives $(a,b) = (10,1)$, which does not work. The former case gives $(a,b) = (6,5)$, which works, and we have $x+y+m = 65 + 56 + 33 = 154\ \mathbf{(E)}$.

Solution 2

The first steps are the same as above. Let $x = 10a+b, y = 10b+a$, where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$. 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 $3^2\cdot 11$. In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even. Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals 11 while $(a-b)$ is a factor of nine (1 or 9). The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$. The sum of the numbers is $\boxed{\textbf{(E) }154}$

Solution 3

Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$. Since $x^2 - y^2 = (x-y)(x+y)$, substituting and factoring, we get $99(a+b)(a-b) = m^2$. Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. Therefore a quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$ Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$. So $m = \sqrt{1089} = 33.$

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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