Difference between revisions of "2005 AMC 12B Problems/Problem 19"
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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. | 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 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{\mathrm{E}}</math> |
== See also == | == See also == | ||
− | + | {{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}} | |
+ | {{MAA Notice}} |
Latest revision as of 19:02, 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.
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 |
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