AMC 8 · 2017 · #3
Grade 8 arithmeticProblem
What is the value of the expression 1684?
Pick an answer.
AMC 8 2017 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: Simplify the nested radical expression $\sqrt{16\sqrt{8\sqrt{4}}}$ and choose the matching answer from (A) $4$, (B) $4\sqrt{2}$, (C) $8$, (D) $8\sqrt{2}$, (E) $16$.
Givens: The expression is a nest of three square roots: $\sqrt{16\sqrt{8\sqrt{4}}}$; The inner constants are $4$, $8$, and $16$ — note that $4$ and $16$ are perfect squares; Multiple-choice options: $4,\;4\sqrt{2},\;8,\;8\sqrt{2},\;16$
Unknowns: The single numerical value of the expression and the matching answer letter
Understand
Restated: Simplify the nested radical expression $\sqrt{16\sqrt{8\sqrt{4}}}$ and choose the matching answer from (A) $4$, (B) $4\sqrt{2}$, (C) $8$, (D) $8\sqrt{2}$, (E) $16$.
Givens: The expression is a nest of three square roots: $\sqrt{16\sqrt{8\sqrt{4}}}$; The inner constants are $4$, $8$, and $16$ — note that $4$ and $16$ are perfect squares; Multiple-choice options: $4,\;4\sqrt{2},\;8,\;8\sqrt{2},\;16$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #11 Work Backwards, #3 Eliminate Possibilities
The expression looks scary because three square roots are stacked, but it splits into three identical sub-tasks: "evaluate one square root, then plug it into the next layer". Tool #7 (Identify Subproblems) turns one big problem into three tiny ones. Tool #11 (Work Backwards) says we must start from the *innermost* root — the outermost layer cannot be touched until the layers beneath it are reduced to plain numbers. Tool #3 (Eliminate Possibilities) is in reserve: the answer choices split cleanly into "integer" and "integer times $\sqrt{2}$" — if every layer produces a perfect square, the result is a clean integer, ruling out (B) and (D).
Execute — Answer: C
8.EE.A.2 Step 1 - Subproblem 1 — peel the innermost square root.
- The deepest layer is $\sqrt{4}$, and $4 = 2 \times 2$, so $\sqrt{4} = 2$.
💡 Knowing that $\sqrt{4} = 2$ uses the square-root symbol's definition — that is the Grade 8 standard for square roots.
3.OA.C.7 Step 2 - Substitute and simplify under the middle radical.
- Replacing $\sqrt{4}$ with $2$, the middle layer becomes $\sqrt{8 \cdot 2} = \sqrt{16}$.
💡 Multiplying $8 \times 2 = 16$ inside the radical is a basic Grade 3 times-table fact.
8.EE.A.2 Step 3 - Subproblem 2 — evaluate the middle square root.
- Since $16 = 4 \times 4$, we have $\sqrt{16} = 4$.
💡 Recognizing $16$ as a perfect square and taking its root is the same Grade 8 square-root standard at work again.
3.OA.C.7 Step 4 - Substitute and simplify under the outermost radical.
- The full expression collapses to $\sqrt{16 \cdot 4} = \sqrt{64}$.
💡 Another single-digit times-table step: $16 \times 4 = 64$, well within Grade 3 fluency.
8.EE.A.2 Step 5 - Subproblem 3 — evaluate the outermost square root.
- Because $64 = 8 \times 8$, the final value is $\sqrt{64} = 8$, which matches choice (C).
💡 $8 \times 8 = 64$ is a perfect square, so the square root is exactly $8$ — Grade 8 square-root symbol use.
8.EE.A.2 Subproblem 1 — peel the innermost square root. The deepest layer is $\sqrt{4}$, 3.OA.C.7 Substitute and simplify under the middle radical. Replacing $\sqrt{4}$ with $2$, 8.EE.A.2 Subproblem 2 — evaluate the middle square root. Since $16 = 4 \times 4$, we have 3.OA.C.7 Substitute and simplify under the outermost radical. The full expression collaps 8.EE.A.2 Subproblem 3 — evaluate the outermost square root. Because $64 = 8 \times 8$, th Review
Reasonableness: Every layer turned into a perfect square ($4$, $16$, $64$), so the answer must be a clean integer — that immediately rules out the $\sqrt{2}$ choices (B) and (D). Among $4$, $8$, and $16$, the result $8$ sits in the middle, which is sensible: nesting roots inside roots tends to shrink large constants ($16$ shrinks toward $\sqrt{16} = 4$), and our calculation gives an answer between the innermost simplification ($\sqrt{4}=2$) and the outer constant ($16$).
Alternative: Tool #3 (Eliminate Possibilities) plus exponent rules: rewrite $\sqrt{16\sqrt{8\sqrt{4}}}$ using $\sqrt{x} = x^{1/2}$. The innermost piece is $4^{1/2}$, the middle layer is $(8 \cdot 4^{1/2})^{1/2} = 8^{1/2} \cdot 4^{1/4} = 2^{3/2} \cdot 2^{1/2} = 2^{2} = 4$, and the outer layer is $(16 \cdot 4)^{1/2} = 64^{1/2} = 8$. Same answer (C), reached by exponent algebra instead of substitution.
CCSS standards used (min grade 8)
3.OA.C.7Fluently multiply and divide within 100 (The under-radical multiplications $8 \times 2 = 16$ and $16 \times 4 = 64$, both single-digit times-table facts within $100$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Evaluating the three square roots $\sqrt{4} = 2$, $\sqrt{16} = 4$, and $\sqrt{64} = 8$ — recognizing each radicand as a perfect square and applying the square-root symbol.)
⭐ This AMC 8 problem only needs the Grade 8 square-root symbol $\sqrt{\;}$ — once you know $\sqrt{4}$, $\sqrt{16}$, and $\sqrt{64}$, the rest is just times tables you already know!
⭐ This AMC 8 problem only needs the Grade 8 square-root symbol $\sqrt{\;}$ — once you know $\sqrt{4}$, $\sqrt{16}$, and $\sqrt{64}$, the rest is just times tables you already know!
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