2009 AMC 10A Problem 13

1. Split $12$ into primes: $12 = 2^2 \times 3$. So
$$12^{mn} = (2^2 \times 3)^{mn} = 2^{2mn} \times 3^{mn}.$$
2. Build $2^{2mn}$ out of $P = 2^m$. Raising $P$ to a power $k$ gives $2^{mk}$. We need $mk = 2mn$, so $k = 2n$:
$$P^{2n} = (2^m)^{2n} = 2^{2mn}.$$
3. Build $3^{mn}$ out of $Q = 3^n$. Raising $Q$ to a power $j$ gives $3^{nj}$. We need $nj = mn$, so $j = m$:
$$Q^{m} = (3^n)^{m} = 3^{mn}.$$
4. Put them together: $12^{mn} = P^{2n} Q^{m}$.
5. Quick check with $m = 1$, $n = 2$: $12^2 = 144$, and $P = 2$, $Q = 9$ give $P^{4} Q^{1} = 16 \times 9 = 144$.

The answer is $\boxed{\textbf{(E)}\ P^{2n}Q^m}$.

Treat each prime on its own. $P$ already carries an $m$ in the exponent of $2$, so it needs one more factor of $2n$ to reach $2mn$. $Q$ carries an $n$ on the $3$, so it needs a factor of $m$. Keeping the primes apart stops you from mixing up which letter goes where.

Plug in small numbers. Take $m = 1$ and $n = 2$, so $12^{mn} = 144$, $P = 2$ and $Q = 9$. Then (A) gives $4 \times 9 = 36$, (B) gives $4 \times 9 = 36$, (C) gives $4 \times 81 = 324$, (D) gives $4 \times 81 = 324$, and (E) gives $16 \times 9 = 144$. Only (E) works. (Be careful: $m = n = 1$ is a poor test, since (A), (D) and (E) all give $12$ then.)