Question:
Suppose $a$ and $b$ are positive integers such that the units digit of $a$ is $2$, the units digit of $b$ is $4$, and the greatest common divisor of $a$ and $b$ is $6$.

What is the smallest possible value of the least common multiple of $a$ and $b$?

Answer:
Both $a$ and $b$ must be divisible by $6$, so the choices for $a$ are $$12, 42, 72, 102, 132, \ldots\phantom{~.}$$and the choices for $b$ are $$24, 54, 84, 114, 144, \ldots~.$$We know that $\mathop{\text{lcm}}[a,b]\cdot \gcd(a,b)=ab$ (since this identity holds for all positive integers $a$ and $b$). Therefore, $$\mathop{\text{lcm}}[a,b] = \frac{ab}{6},$$so in order to minimize $\mathop{\text{lcm}}[a,b]$, we should make $ab$ as small as possible. But we can't take $a=12$ and $b=24$, because then $\gcd(a,b)$ would be $12$, not $6$. The next best choice is either $a=12,b=54$ or $a=42,b=24$. Either of these pairs yields $\gcd(a,b)=6$ as desired, but the first choice, $a=12$ and $b=54$, yields a smaller product. Hence this is the optimal choice, and the smallest possible value for $\mathop{\text{lcm}}[a,b]$ is $$\mathop{\text{lcm}}[12,54] = \frac{12\cdot 54}{6} = 2\cdot 54 = \boxed{108}.$$