2018 AMC 10B Problem 23

Let $g = \gcd(a, b)$ and write $a = gx$, $b = gy$ with $\gcd(x, y) = 1$. Then $\operatorname{lcm}(a, b) = gxy$ and $ab = g^2 xy$. The equation becomes
$$g^2 xy + 63 = 20 gxy + 12 g \quad\Longrightarrow\quad 63 = g\big(20xy + 12 - gxy\big) = g\big(xy(20 - g) + 12\big).$$
So $g$ is a divisor of $63$: $g \in \{1, 3, 7, 9, 21, 63\}$, and $\frac{63}{g} - 12 = xy(20 - g)$.

• $g = 1$: $51 = 19xy$, no.
• $g = 3$: $9 = 17xy$, no.
• $g = 7$: $-3 = 13xy$, no.
• $g = 9$: $-5 = 11xy$, no.
• $g = 21$: $-9 = -xy$, so $xy = 9$ with $x, y$ coprime: $(x, y) = (1, 9)$ or $(9, 1)$.
• $g = 63$: $-11 = -43xy$, no.

The coprime factorizations of $9$ exclude $(3, 3)$, leaving two ordered pairs: $(a, b) = (21, 189)$ and $(189, 21)$. Check: $21 \cdot 189 + 63 = 4032$ and $20 \cdot 189 + 12 \cdot 21 = 3780 + 252 = 4032$.

The answer is $\boxed{\textbf{(B)}\ 2}$.

Any equation mixing $\gcd$, $\operatorname{lcm}$ and the product should be rewritten with $a = gx$, $b = gy$, $\gcd(x,y) = 1$, because then $\operatorname{lcm} = gxy$ and everything becomes polynomial. Isolating a constant as $g \cdot (\ldots)$ turns an infinite search into casework on the divisors of that constant.

Use $ab = \gcd \cdot \operatorname{lcm}$ directly: with $G = \gcd$, $L = \operatorname{lcm}$, the equation is $GL - 20L - 12G + 63 = 0$, i.e. $(G - 20)(L - 12) = 240 - 63 = 177 = 3 \cdot 59$ (Simon's trick). Since $G \mid L$ and $L \ge G$, test the factor pairs: $G - 20 = 1, L - 12 = 177$ gives $(21, 189)$ and works; the others fail divisibility or positivity. Then $L/G = 9$ splits into coprime parts as $1 \cdot 9$ in two orders.

• Forgetting that x and y must be coprime, so that xy = 9 wrongly admits (3,3) (which has gcd 63, not 21) and yields 3 ordered pairs instead of 2.
• Counting $(21, 189)$ once and answering with an unordered count, or forgetting to reject the cases where $xy(20 - g)$ has the wrong sign.