Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 19: Line 19:
 
<cmath>=(11(a+b))(9(a-b))</cmath>
 
<cmath>=(11(a+b))(9(a-b))</cmath>
  
−
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 9 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{E}</math>
+
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{E}</math>
  
 
== See also ==
 
== See also ==
 
* [[2005 AMC 12B Problems]]
 
* [[2005 AMC 12B Problems]]

Revision as of 11:24, 4 February 2011

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 $x+y+m=154\Rightarrow\boxed{E}$

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_19&oldid=36567"