Difference between revisions of "2005 AMC 12B Problems/Problem 19"
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<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 10:24, 4 February 2011
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.
By difference of squares,
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