AMC 10 · 2009 · #6

Grade 8 algebra
exponentsprime-factorization convert-to-algebra ↑ Prerequisites: exponents
📏 Medium solution 💡 2 insights
Problem
Two letters stand for powers of two different primes. Find which expression always equals a given power of twelve.

Pick an answer.

(A)
$P^2Q$
(B)
$P^nQ^m$
(C)
$P^nQ^{2m}$
(D)
$P^{2m}Q^n$
(E)
$P^{2n}Q^m$
How to solve
Strategy Introduce a Variable

Everything is written with the two building blocks P = 2^m and Q = 3ⁿ. Break the base 12 into its prime powers of 2 and 3, then rebuild each prime power out of P and Q using the exponent rules. A quick number test at the end pins down the single matching choice.

1STEP 1

Factor the base into primes

Factoring the base gives two prime powers.

12^mn = (2² · 3)^mn = 2²mn · 3^mn
2STEP 2

Rebuild the prime powers from P and Q

One prime appears twice, so its exponent doubles.

P²ⁿ = 2²mn, Q^m = 3^mn
3STEP 3

Match the pieces together

Matching the pieces gives P²ⁿQ^m.

12^mn = 2²mn · 3^mn = P²ⁿ Q^m
4STEP 4

Confirm with a quick number test

A quick numeric test confirms P²ⁿQ^m, choice (D).

m=2, n=1: 12^mn=144 = 2⁴ · 3 = P²ⁿQ^m
Answer
P²ⁿQ^m
The base 12 = 2² · 3 carries twice as many factors of 2 as of 3, so the 2-part should have the bigger exponent. Choice (E) uses P²ⁿ (the 2-part) against Q^m (the 3-part), and 2n beats m in exactly the doubled way 2² · 3 predicts. Testing m=2, n=1 gives 144 on both sides, confirming the match holds beyond a single lucky pair.
💡Key takeaway

Break a number into its prime powers, then rebuild the target one prime at a time using the exponent rules.

  • Factor the base into primes
  • Rebuild the prime powers from P and Q
  • Match the pieces together
  • Confirm with a quick number test