# Difference between revisions of "2018 AMC 10B Problems/Problem 23"

(Work in progress of my answer to this question.) |
(This is my solution to the problem.) |
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+ | ==Solution 1== | ||

Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes | Let <math>x = lcm(a, b)</math>, and <math>y = gcd(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes | ||

− | <cmath>x\cdot y + 63 = 20x + 12y</cmath> | + | <cmath>x\cdot y + 63 = 20x + 12y</cmath> |

− | <cmath>x\cdot y - 20x - 12y + 63 = 0</cmath> | + | <cmath>x\cdot y - 20x - 12y + 63 = 0</cmath> |

Using Simon's Favorite Factoring Trick, we rewrite this equation as | Using Simon's Favorite Factoring Trick, we rewrite this equation as | ||

<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath> | <cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath> | ||

− | <cmath>(x - 12)(y - 20) = 177</cmath>. | + | <cmath>(x - 12)(y - 20) = 177</cmath> |

+ | |||

+ | Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>, which can be translated back to two solution for <math>a</math> and <math>b</math>. Thus, the answer is <math>\boxed{2}</math>. | ||

(awesomeag) | (awesomeag) |

## Revision as of 15:34, 16 February 2018

23. How many ordered pairs of positive integers satisfy the equation where denotes the greatest common divisor of and , and denotes their least common multiple?

## Solution 1

Let , and . Therefore, . Thus, the equation becomes

Using Simon's Favorite Factoring Trick, we rewrite this equation as

Since and , we have and , or and . This gives us the solutions and , which can be translated back to two solution for and . Thus, the answer is . (awesomeag)