AMC 8 · 2011 · #17

Easy mode Grade 6

Problem

The symbols $2^w$ , $3^x$ , $5^y$ , $7^z$ 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 ( $0$ , $1$ , $2$ , $3$ , $\ldots$ ). Remember that any number to the $0$ power equals $1$ .

Suppose $w, x, y, z$ are picked so that
$2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588.$

Once you find $w$ , $x$ , $y$ , and $z$ , what is $2w + 3x + 5y + 7z$ ?

$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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 $2w + 3x + 5y + 7z$.

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

#7 Identify Subproblems 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 $49 = 7 \cdot 7$.

$$588 = 2 \cdot 2 \cdot 3 \cdot 7 \cdot 7 = 2^2 \cdot 3^1 \cdot 7^2$$

💡 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.

#11 Work Backwards 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 $1$.

$$w = 2, \;\; x = 1, \;\; y = 0, \;\; z = 2$$

💡 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.

#7 Identify Subproblems 5.OA.A.1 Step 3

Plug the four values into the target expression and compute.

$$2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2) = 4 + 3 + 0 + 14 = 21 \;\Rightarrow\; \textbf{(A)}$$

💡 Evaluating a numerical expression with multiplication and addition follows Grade 5 order-of-operations standard 5.OA.A.1.

[1] #7 4.OA.B.4 Subproblem 1: factor $588$ into primes. Divide by the smallest prime that fits,

[2] #11 6.EE.A.1 Subproblem 2: match exponents. Working backwards (Tool #11) from $2^w \cdot 3^x

[3] #7 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.4 Find 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.1 Write 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.1 Use 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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