The greatest common divisor of positive integers $m$ and $n$ is 6. The least common multiple of $m$ and $n$ is 126. What is the least possible value of $m+n$?
To find the least possible value of $m+n$, I want to minimize both $m$ and $n$. I know that the greatest common divisor of $m$ and $n$ is 6, which means that both $m$ and $n$ are multiples of 6. So I can write $m=6a$ and $n=6b$ for some positive integers $a$ and $b$. Now I can use the formula for the least common multiple: $LCM(m,n)=\frac{m\cdot n}{GCD(m,n)}$. Plugging in the given values, I get: $126=\frac{6a\cdot 6b}{6}$. Simplifying, I get: $126=6ab$. Dividing both sides by 6, I get: $21=ab$. Now I want to find the smallest possible values of $a$ and $b$ that satisfy this equation. Since $a$ and $b$ are positive integers, they must be factors of 21. The factors of 21 are 1, 3, 7, and 21. If I choose $a=1$ and $b=21$, then $m+n=6a+6b=6+126=132$. But this is not the smallest possible value, because I can choose smaller values for $a$ and $b$. If I choose $a=3$ and $b=7$, then $m+n=6a+6b=18+42=60$. This is smaller than 132, but can I do better? The only other pair of factors of 21 is $a=7$ and $b=3$, but this gives the same value of $m+n$ as before. So the smallest possible value of $m+n$ is 60. 