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 16:34, 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)