Question

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

$\begin{aligned} a \cdot b + 63 &= 20 \cdot \operatorname{lcm}(a, b) \\ &\quad {}+ 12 \cdot \gcd(a, b), \end{aligned}$

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

$0$

$2$

$4$

$6$

$8$

Answer: B

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

Difficulty rating: 2120

Solution:

Recall $ab=\gcd(a,b)\operatorname{lcm}(a,b).$ Let $x=\operatorname{lcm}(a,b)$ and $y=\gcd(a,b).$ The equation becomes $(x-12)(y-20)=177=3\cdot59.$

The positive factor pairs give $(x,y)=(13,197),$ $(189,21),$ $(15,79),$ and $(71,23).$ The negative factor pairs make either $x$ or $y$ negative, so they are impossible. Also $y$ must divide $x,$ and only $(x,y)=(189,21)$ passes.

Write $a=21u$ and $b=21v.$ Then $\gcd(u,v)=1$ and $uv=189/21=9,$ so $(u,v)=(1,9)$ or $(9,1).$ Hence the two ordered pairs are $(21,189)$ and $(189,21),$ and B is the correct answer.

2018 AMC 10B Problem 23

Answer: B

Difficulty rating: 2120

Solution:

Problem 23 in Other Years