AMC 8 · 2011 · #17
Easy mode Grade 6Problem
The symbols 2w, 3x, 5y, 7z mean 2 multiplied by itself w times, 3 multiplied by itself x times, and so on. The letters w,x,y,z stand for whole numbers (0, 1, 2, 3, …). Remember that any number to the 0 power equals 1.
Suppose w,x,y,z are picked so that
2w⋅3x⋅5y⋅7z=588.
Once you find w, x, y, and z, what is 2w+3x+5y+7z?
Pick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Whole numbers $w$, $x$, $y$, $z$ satisfy $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$. Find $2w + 3x + 5y + 7z$.
Givens: $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$; $w$, $x$, $y$, $z$ are whole numbers (so $0$ is allowed); The four bases $2, 3, 5, 7$ are all prime; Answer choices: (A) $21$, (B) $25$, (C) $27$, (D) $35$, (E) $56$
Unknowns: The value of the linear combination $2w + 3x + 5y + 7z$
Understand
Restated: Whole numbers $w$, $x$, $y$, $z$ satisfy $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$. Find $2w + 3x + 5y + 7z$.
Givens: $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$; $w$, $x$, $y$, $z$ are whole numbers (so $0$ is allowed); The four bases $2, 3, 5, 7$ are all prime; Answer choices: (A) $21$, (B) $25$, (C) $27$, (D) $35$, (E) $56$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #11 Work Backwards
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.
Execute — Answer: A
4.OA.B.4 Step 1 - Subproblem 1: factor $588$ into primes.
- Divide by the smallest prime that fits, repeatedly.
- $588$ is even, so pull out $2$ twice: $588 = 2 \cdot 294 = 2 \cdot 2 \cdot 147$.
- Then $147$ has digit sum $1+4+7=12$, divisible by $3$, so $147 = 3 \cdot 49$.
- Finally $49 = 7 \cdot 7$.
💡 Finding factor pairs and identifying prime factors is exactly the Grade 4 standard 4.OA.B.4 — recognize that whole numbers are products of primes.
6.EE.A.1 Step 2 - Subproblem 2: match exponents.
- Working backwards (Tool #11) from $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 2^2 \cdot 3^1 \cdot 7^2$, read off each exponent.
- The factor $5$ does not appear in $588$, so $5^y = 1$, which forces $y = 0$ because any nonzero base to the $0$ is $1$.
💡 Reading off exponents from a factorization is the Grade 6 "whole-number exponents" standard 6.EE.A.1 in reverse — you know the value, you recover the exponent.
5.OA.A.1 Step 3 Plug the four values into the target expression and compute.
💡 Evaluating a numerical expression with multiplication and addition follows Grade 5 order-of-operations standard 5.OA.A.1.
4.OA.B.4 Subproblem 1: factor $588$ into primes. Divide by the smallest prime that fits, 6.EE.A.1 Subproblem 2: match exponents. Working backwards (Tool #11) from $2^w \cdot 3^x 5.OA.A.1 Plug the four values into the target expression and compute. Review
Reasonableness: Sanity-check the factorization: $2^2 \cdot 3 \cdot 7^2 = 4 \cdot 3 \cdot 49 = 12 \cdot 49 = 588$. ✓ 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.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: notice $5y$ is a multiple of $5$ and $2w$ is even. Trying $y = 0$ (since $588$ is clearly not a multiple of $5$ — it doesn't end in $0$ or $5$) immediately kills the $5y$ term. Then any answer must equal $2w + 3x + 7z$ with small exponents; only $21 = 4 + 3 + 14$ matches a clean factorization of $588$, confirming (A).
CCSS standards used (min grade 6)
4.OA.B.4Find factor pairs and recognize prime/composite numbers (Breaking $588$ down into its prime factors $2 \cdot 2 \cdot 3 \cdot 7 \cdot 7$ by repeated division by the smallest primes.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Reading $2^2 \cdot 3^1 \cdot 7^2$ as the unique exponent form of $588$ and recovering $w = 2, x = 1, y = 0, z = 2$.)5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate them (Evaluating $2(2) + 3(1) + 5(0) + 7(2) = 21$ with the correct order of operations.)
⭐ This AMC 8 problem only needs Grade 6 exponent reasoning — break $588$ into primes, read off the exponents, plug in — that you already know!
⭐ 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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