2009 AMC 10A Problem 13

Since $12 = 2^2 \cdot 3$,
$$12^{mn} = (2^2 \cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn}.$$

Now express each factor using $P = 2^m$ and $Q = 3^n$. To turn $2^m$ into $2^{2mn}$ raise it to the power $2n$: $P^{2n} = 2^{2mn}$. To turn $3^n$ into $3^{mn}$ raise it to the power $m$: $Q^m = 3^{mn}$.

Therefore $12^{mn} = P^{2n} Q^m$.

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

Everything reduces to matching exponents prime by prime. $P$ already carries the exponent $m$ on the prime $2$, so the extra factor it needs is whatever is left of $2mn$, namely $2n$; likewise $Q$ carries $n$ on the prime $3$ and needs $m$. Keeping the primes separate prevents mixing up which exponent belongs to which base.

Plug in $m = 1$, $n = 2$: then $12^{2} = 144$, $P = 2$, $Q = 9$. Only choice (E) gives $2^4 \cdot 9 = 144$; for instance (D) gives $2^2 \cdot 9 = 36$. (Using $m = n = 1$ is not enough, since (A), (D) and (E) all give $12$.)

• Attaching the exponent to the wrong variable, e.g. 2^(2mn) = (2^m)^(2m), and picking (D) P^(2m) Q^n.
• Testing only $m = n = 1$, which fails to separate (A), (D) and (E).