Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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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 | ||
Revision as of 16:58, 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
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 . Obviously, the first pair does not work. The second pair can be ordered in two ways. Thus, the answer is . (awesomeag)
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
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