Difference between revisions of "2018 AMC 10B Problems/Problem 23"
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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>. | 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) | ||
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+ | ==See Also== | ||
+ | {{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Revision as of 17:12, 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)
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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