AMC 8 · 2011 · #17

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

prime-factorizationexponents identify-subproblems ↑ Prerequisites: prime-factorization

📏 Medium solution 💡 3 insights

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Problem

Whole numbers w, x, y, z satisfy 2^w · 3^x · 5^y · 7^z = 588. Find 2w + 3x + 5y + 7z.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The expression 2w + 3x + 5y + 7z depends on four unknown exponents, so the problem really has two clean subproblems (Tool #7): (1) find the prime factorization of 588, and (2) read off w, x, y, z by matching exponents, then plug in. Tool #11 (Work Backwards) captures the reverse-engineering move: we are handed the product 588 and have to recover the exponents that built it. We deliberately avoid Tool #13 (Algebra) because no equation manipulation is needed — pulling out prime factors and matching is enough.

1STEP 1

Pull out 2 twice from 588, then 3 from 147, leaving 49 = 7 · 7, so 588 = 2² · 3¹ · 7².

588 = 2 · 2 · 3 · 7 · 7 = 2² · 3¹ · 7²

2STEP 2

Match same bases to read the exponents; base 5 is absent so y = 0, giving w = 2, x = 1, z = 2.

w = 2, x = 1, y = 0, z = 2

3STEP 3

Substitute w = 2, x = 1, y = 0, z = 2 into 2w + 3x + 5y + 7z to get 21.

2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21 → (A)

Answer

Sanity-check the factorization: 2² · 3 · 7² = 4 · 3 · 49 = 12 · 49 = 588. ✓ The exponents are all small whole numbers, as expected for an AMC 8 problem. The final value 21 is the smallest of the five answer choices, which fits — most exponents are tiny (0, 1, 2), so the linear combination should be modest.

💡Key takeaway

This AMC 8 problem only needs Grade 6 exponent reasoning — break 588 into primes, read off the exponents, plug in — that you already know!

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