2018 AMC 10B Problem 23

An equation with gcd and lcm is easiest after pulling the gcd out of both numbers.

1. Pull out the gcd. Let $g = \gcd(a, b)$. Then $a = gx$ and $b = gy$, where $x$ and $y$ share no common factor (otherwise $g$ would not be the greatest). For example, $a = 12$, $b = 18$ gives $g = 6$, $x = 2$, $y = 3$.
2. Rewrite each piece.
- $a \cdot b = g^2 xy$.
- $\text{lcm}(a, b) = gxy$. (In the example, $6 \times 2 \times 3 = 36$.)
3. Substitute.
$$g^2 xy + 63 = 20gxy + 12g.$$
Move the $xy$ terms to one side:
$$63 = 20gxy - g^2xy + 12g = g\big(xy(20 - g) + 12\big).$$
4. So $g$ divides $63$. The right side is $g$ times a whole number, so $g$ is one of $1, 3, 7, 9, 21, 63$. Dividing by $g$:
$$\frac{63}{g} - 12 = xy(20 - g).$$
5. Check each $g$. Remember $xy$ must be a positive whole number.

6. Only $g = 21$ works, with $xy = 9$. The ways to write $9$ as $x \times y$ are $1 \times 9$, $9 \times 1$ and $3 \times 3$. But $3$ and $3$ share the factor $3$, so that one is out. That leaves $(x, y) = (1, 9)$ or $(9, 1)$, giving
$$(a, b) = (21, 189) \text{ or } (189, 21).$$
7. Check. $21 \times 189 + 63 = 3969 + 63 = 4032$, and $20 \times 189 + 12 \times 21 = 3780 + 252 = 4032$.

There are $2$ ordered pairs.

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

Writing $a = gx$, $b = gy$ with $x$ and $y$ sharing no factor turns gcd and lcm into ordinary multiplication. Then pulling $g$ out of the equation shows $g$ must divide a fixed number, so only a handful of cases need checking. Always remember the "no shared factor" rule when you rebuild $a$ and $b$.

Use the fact that $a \cdot b = \gcd \times \text{lcm}$. Write $G$ for the gcd and $L$ for the lcm. The equation becomes $GL + 63 = 20L + 12G$, which rearranges to
$$(G - 20)(L - 12) = 240 - 63 = 177 = 3 \times 59.$$
Try the factor pairs. $G - 20 = 1$ and $L - 12 = 177$ give $G = 21$, $L = 189$, and $189 = 21 \times 9$, so it works. $G = 23$, $L = 71$ fails because $71$ is not a multiple of $23$. The other pairs make $L$ smaller than $G$ or make a value negative. With $G = 21$ and $L = 189$, the two numbers are $21$ and $189$ in either order.

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 $\boxed{\textbf{(B)}\ 2}$.