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2018 AMC 10B Problem 23

Problem 23 of 25HarderAlgebraNumber Theory

How many ordered pairs (a,b)(a, b) of positive integers satisfy the equation ab+63=20lcm(a,b)+12gcd(a,b), \begin{aligned} a \cdot b + 63 &= 20 \cdot \operatorname{lcm}(a, b) \\ &\quad {}+ 12 \cdot \gcd(a, b), \end{aligned} where gcd(a,b)\gcd(a, b) denotes the greatest common divisor of aa and b,b, and lcm(a,b)\operatorname{lcm}(a, b) denotes their least common multiple?

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Solution

Recall ab=gcd(a,b)lcm(a,b).ab=\gcd(a,b)\operatorname{lcm}(a,b). Let x=lcm(a,b)x=\operatorname{lcm}(a,b) and y=gcd(a,b).y=\gcd(a,b). The equation becomes (x12)(y20)=177=359.(x-12)(y-20)=177=3\cdot59. The positive factor pairs give (x,y)=(13,197),(x,y)=(13,197), (189,21),(189,21), (15,79),(15,79), and (71,23).(71,23). The negative factor pairs make either xx or yy negative, so they are impossible. Also yy must divide x,x, and only (x,y)=(189,21)(x,y)=(189,21) passes. Write a=21ua=21u and b=21v.b=21v. Then gcd(u,v)=1\gcd(u,v)=1 and uv=18921=9,uv=\frac{189}{21}=9, so (u,v)=(1,9)(u,v)=(1,9) or (9,1).(9,1). Hence the two ordered pairs are (21,189)(21,189) and (189,21),(189,21), and B is the correct answer.

More practice

Concepts: least common multiple · greatest common divisor · Simon’s Favorite Factoring Trick

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.