Competition · AMC preparation · step 4 of 4

AMC 8 · 2017 · #3

Grade 8 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
perfect-squaresexponentsorder-of-operations identify-subproblems ↑ Prerequisites: multi-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
Simplify the nested radical expression 16√(8√(4)) and choose the matching answer from (A) 4, (B) 4√(2), (C) 8, (D) 8√(2), (E) 16.

Pick an answer.

(A)
4
(B)
$4\sqrt{2}$
(C)
8
(D)
$8\sqrt{2}$
(E)
16

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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 √(2)" — if every layer produces a perfect square, the result is a clean integer, ruling out (B) and (D).

1STEP 1

Take the innermost root

Peel the innermost root first: since 4 = 2 × 2, we get √(4) = 2.

√(4) = 2
2STEP 2

Substitute under the middle root

Substitute 2 into the middle layer: √(8 · 2) = √(16).

√(8 √(4)) = √(8 · 2) = √(16)
3STEP 3

Evaluate the middle root

Evaluate the middle root: since 16 = 4 × 4, √(16) = 4.

√(16) = 4
4STEP 4

Substitute under the outer root

Substitute 4 into the outer layer: the whole expression collapses to √(16 · 4) = √(64).

√(16 √(8 √(4))) = √(16 · 4) = √(64)
5STEP 5

Evaluate the outer root

Evaluate the outer root: since 64 = 8 × 8, the final value is √(64) = 8, choice (C).

√(64) = 8 → (C)
Answer
8
Every layer turned into a perfect square (4, 16, 64), so the answer must be a clean integer — that immediately rules out the √(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 √(16) = 4), and our calculation gives an answer between the innermost simplification (√(4)=2) and the outer constant (16).
💡Key takeaway

This AMC 8 problem only needs the Grade 8 square-root symbol √( ) — once you know √(4), √(16), and √(64), the rest is just times tables you already know!

  • Take the innermost root
  • Substitute under the middle root
  • Evaluate the middle root
  • Substitute under the outer root
  • Evaluate the outer root

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