If $m$ and $n$ are positive integers such that $\gcd(m,n) = 12$, then what is the smallest possible value of $\gcd(10m,15n)$?
Since $\gcd(m,n) = 12$, both $m$ and $n$ are divisible by 12.  Then $10m$ is divisible by $10 \cdot 12 = 120$, and $15n$ is divisible by $12 \cdot 15 = 180$.  Since 60 divides both 120 and 180, $\gcd(10m,15n)$ must be at least 60.

If we set $m = n = 12$, then $\gcd(m,n) = \gcd(12,12) = 12$, and $\gcd(10m,15n) = \gcd(120,180) = 60$, which shows that the value of 60 is attainable.  Therefore, the smallest possible value of $\gcd(10m,15n)$ is $\boxed{60}$.
60
Level 4
Number Theory
Problem: If $m$ and $n$ are positive integers such that $\gcd(m,n) = 12$, then what is the smallest possible value of $\gcd(10m,15n)$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the smallest possible value of $\gcd(10m, 15n)$ given that $\gcd(m, n) = 12$, we start by expressing $m$ and $n$ in terms of their greatest common divisor. Let $m = 12k$ and $n = 12j$ for some positive integers $k$ and $j$ such that $\gcd(k, j) = 1$. 

Next, we compute $10m$ and $15n$:
10m = 10(12k) = 120k
15n = 15(12j) = 180j

Now, we need to find $\gcd(120k, 180j)$. We can factor out the common terms:
gcd(120k, 180j) = gcd(120, 180) \cdot \gcd(k, j)

Calculating $\gcd(120, 180)$:
120 = 2^3 * 3 * 5
180 = 2^2 * 3^2 * 5

The minimum power of each prime factor gives:
gcd(120, 180) = 2^2 * 3^1 * 5^1 = 60

Since $\gcd(k, j) = 1$, we have:
gcd(120k, 180j) = 60 * 1 = 60

Thus, the smallest possible value of $\gcd(10m, 15n)$ is 60.<|END_OF_LATENT|> The final answer is $60$.