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Difference between revisions of "2005 AMC 12B Problems/Problem 19"

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== Solution ==
 
== Solution ==
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let <math>x=10a+b</math>, then <math>y=10b+a</math> where <math>a</math> and <math>b</math> are nonzero digits.
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By [[difference of squares]],
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<cmath>x^2-y^2=(x+y)(x-y)</cmath>
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<cmath>=(10a+b+10b+a)(10a+b-10b-a)</cmath>
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<cmath>=(11(a+b))(9(a-b))</cmath>
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For this product to be a square, the factor of <math>11</math> must be repeated in either <math>(a+b)</math> or <math>(a-b)</math>, and given the constraints it has to be <math>(a+b)=11</math>. The factor of <math>9</math> is already a square and can be ignored. Now <math>(a-b)</math> must be another square, and since <math>a</math> cannot be <math>10</math> or greater then <math>(a-b)</math> must equal <math>4</math> or <math>1</math>. If <math>a-b=4</math> then <math>(a+b)+(a-b)=11+4</math>, <math>2a=15</math>, <math>a=15/2</math>, which is not a digit. Hence the only possible value for <math>a-b</math> is <math>1</math>. Now we have <math>(a+b)+(a-b)=11+1</math>, <math>2a=12</math>, <math>a=6</math>, then <math>b=5</math>, <math>x=65</math>, <math>y=56</math>, <math>m=33</math>, and <math>x+y+m=154\Rightarrow\boxed{\mathrm{E}}</math>
  
 
== See also ==
 
== See also ==
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* [[2005 AMC 12B Problems]]
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{{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 19:02, 24 December 2020

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 $x+y+m$?

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

Solution

let $x=10a+b$, then $y=10b+a$ where $a$ and $b$ are nonzero digits.

By difference of squares, \[x^2-y^2=(x+y)(x-y)\] \[=(10a+b+10b+a)(10a+b-10b-a)\] \[=(11(a+b))(9(a-b))\]

For this product to be a square, the factor of $11$ must be repeated in either $(a+b)$ or $(a-b)$, and given the constraints it has to be $(a+b)=11$. The factor of $9$ is already a square and can be ignored. Now $(a-b)$ must be another square, and since $a$ cannot be $10$ or greater then $(a-b)$ must equal $4$ or $1$. If $a-b=4$ then $(a+b)+(a-b)=11+4$, $2a=15$, $a=15/2$, which is not a digit. Hence the only possible value for $a-b$ is $1$. Now we have $(a+b)+(a-b)=11+1$, $2a=12$, $a=6$, then $b=5$, $x=65$, $y=56$, $m=33$, and $x+y+m=154\Rightarrow\boxed{\mathrm{E}}$

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 12 Problems and Solutions

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