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

 
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== Problem ==
 
== Problem ==
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Let <math>x</math> and <math>y</math> be two-digit integers such that <math>y</math> is obtained by reversing the digits
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of <math>x</math>. The integers <math>x</math> and <math>y</math> satisfy <math>x^2 - y^2 = m^2</math> for some positive integer <math>m</math>.
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What is <math>x + y + m</math>?
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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>
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== Solution ==
 
== Solution ==
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Let <math>x = 10a+b, y = 10b+a</math>, [[without loss of generality]] with <math>a>b</math>. Then <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. It follows that <math>11|(a-b)(a+b)</math>, but <math>a-b < 10</math> so <math>11|a+b \Longrightarrow a+b=11</math>. Then we have <math>33^2(a-b) = m^2</math>. Thus <math>a-b</math> is a perfect square. Also, since <math>a-b</math> and <math>a+b</math> have the same parity, so <math>a-b</math> is a one-digit odd perfect square, namely <math>1</math> or <math>9</math>. The latter case gives <math>(a,b) = (10,1)</math>, which does not work. The former case gives <math>(a,b) = (6,5)</math>, which works, and we have <math>x+y+m = 65 + 56 + 33 = 154\ \mathbf{(E)}</math>.
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== See Also ==
 
== See Also ==
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*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|n-b=23|n-a=25}}
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[[Category:Introductory Number Theory Problems]]

Revision as of 19:52, 2 January 2009

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, since $a-b$ and $a+b$ have the same parity, so $a-b$ 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 $x+y+m = 65 + 56 + 33 = 154\ \mathbf{(E)}$.

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions
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