AMC 8 · 2018 · #25

Grade 8 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
exponentsperfect-squaresbound-inequality-then-enumerate bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: exponentsmulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
Count every perfect cube n³ (where n is a positive integer) that lies in the closed interval [2⁸+1, 2¹⁸+1]. The endpoints themselves count if they happen to be perfect cubes.

Pick an answer.

(A)
4
(B)
9
(C)
10
(D)
57
(E)
58

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

"How many cubes lie in [L, U]?" naturally splits (Tool #7) into three subproblems: (a) find the smallest integer n_min with n_min³ ≥ L, (b) find the largest integer n_max with n_max³ ≤ U, (c) count the integers from n_min to n_max inclusive. For (a) the lower bound 257 is small enough that Tool #6 (Guess & Check on 6³, 7³) lands instantly. For (b) the upper bound 2¹⁸+1 looks scary, but Tool #9 (rewrite to an easier form) reveals the trick: 18 = 6 × 3, so 2¹⁸ = (2⁶)³ = 64³ — a perfect cube already sitting just below the upper bound.

1STEP 1

Subproblem A: compute 2⁸+1 = 257, then guess-and-check small cubes — 6³ = 216 is too small, 7³ = 343 clears it.

6³ = 216 ＜ 257, 7³ = 343 ≥ 257
2STEP 2

The smallest cube in range is 7³ = 343; any base ≤ 6 falls below 257, so n_min = 7.

n_min = 7
3STEP 3

Subproblem B: instead of computing 2¹⁸, use 18 = 6×3 to write 2¹⁸ = (2⁶)³ = 64³, so the upper bound is 64³+1.

2¹⁸ = (2⁶)³ = 64³, upper bound = 64³ + 1
4STEP 4

Compare: 64³ ≤ 64³+1 (included) but 65³ ＞ 64³+1 (excluded), so the largest base is n_max = 64.

n_max = 64
5STEP 5

Subproblem C: count integers n = 7, 8, …, 64. Inclusive count is 64 − 7 + 1 = 58, matching choice (E).

64 - 7 + 1 = 58 → (E)
Answer
58
The smallest cube is 7³ and the largest is 64³. The count is 64 - 7 + 1 = 58, which is exactly choice (E). A quick sanity check on the "off-by-one": from 7 to 64 there are clearly more than 50 integers (since 64 - 7 = 57 gaps, plus one for the inclusive endpoint), ruling out (A)=4, (B)=9, (C)=10. The remaining trap is (D)=57, which is the mistake of forgetting the +1 — easy to spot since the problem says "inclusive".
💡Key takeaway

This AMC 8 problem only needs Grade 8 exponent rules — like (2⁶)³ = 2¹⁸ — you already know!

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