Difference between revisions of "2005 AMC 12B Problems/Problem 19"
Arthy00009 (talk | contribs) (→Solution) |
|||
| Line 14: | Line 14: | ||
let <math>x=10a+b</math>, then <math>y=10b+a</math> where <math>a</math> and <math>b</math> are nonzero digits. | let <math>x=10a+b</math>, then <math>y=10b+a</math> where <math>a</math> and <math>b</math> are nonzero digits. | ||
| − | By difference of squares, | + | By [[difference of squares]], |
<cmath>x^2-y^2=(x+y)(x-y)</cmath> | <cmath>x^2-y^2=(x+y)(x-y)</cmath> | ||
<cmath>=(10a+b+10b+a)(10a+b-10b-a)</cmath> | <cmath>=(10a+b+10b+a)(10a+b-10b-a)</cmath> | ||
<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 <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> | + | 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 == | ||
{{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:02, 24 December 2020
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
let 
, then 
where 
and 
are nonzero digits.
![\[x^2-y^2=(x+y)(x-y)\]](http://latex.artofproblemsolving.com/1/b/8/1b8088e58252db3fe978b651bf66d791f79fb190.png)
![\[=(10a+b+10b+a)(10a+b-10b-a)\]](http://latex.artofproblemsolving.com/f/7/5/f75db726bece8fe4b6cf1036ff0d80353d418f83.png)
![\[=(11(a+b))(9(a-b))\]](http://latex.artofproblemsolving.com/4/3/d/43d794211b54da0f1669ac78864b06348267c4fb.png)
For this product to be a square, the factor of 
must be repeated in either 
or 
, and given the constraints it has to be 
. The factor of 
is already a square and can be ignored. Now must be another square, and since cannot be 
or greater then must equal 
or 
. If 
then 
, 
, 
, which is not a digit. Hence the only possible value for 
is . Now we have 
, 
, 
, then 
, 
, 
, 
, and 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
