Question

How many ordered pairs $(a, b)$ of positive integers satisfy the equation

$a \cdot b + 63 = 20 \cdot \operatorname{lcm}(a, b) + 12 \cdot \gcd(a, b),$

where $\gcd(a, b)$ denotes the greatest common divisor of $a$ and $b,$ and $\operatorname{lcm}(a, b)$ denotes their least common multiple?

$0$

$2$

$4$

$6$

$8$

Solution:

Recall $ab = \gcd(a,b) \cdot \operatorname{lcm}(a,b).$ Let $x = \operatorname{lcm}(a,b)$ and $y = \gcd(a,b).$ The equation turns into $xy + 63 = 20x + 12y,$ which factors as $(x - 12)(y - 20) = 177 = 3 \cdot 59.$ The positive factorizations give $(x, y) = (13, 197), (189, 21), (15, 79), (71, 23).$ But we also need $y \mid x,$ and only $(189, 21)$ passes. So $\gcd = 21,$ $\operatorname{lcm} = 189,$ and $\{a, b\} = \{21, 189\}.$ That's the $2$ ordered pairs $(21, 189)$ and $(189, 21).$ Thus, B is the correct answer.

2018 AMC 10B Problem 23

Solution:

Problem 23 in Other Years