Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #25
Grade 8 number-theoryPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
"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.
Find the smallest cube
Subproblem A: compute 2⁸+1 = 257, then guess-and-check small cubes — 6³ = 216 is too small, 7³ = 343 clears it.
Evaluating 2⁸ and small whole-number cubes uses whole-number exponent expressions — a Grade 6 skill.
6.EE.A.1Guess And CheckSet the lower index
The smallest cube in range is 7³ = 343; any base ≤ 6 falls below 257, so n_min = 7.
Choosing n_min = 7 is finding the smallest n that makes the inequality n³ ≥ 257 true — a Grade 6 "solve an inequality by testing" move.
6.EE.B.5Identify SubproblemsRewrite the upper bound
Subproblem B: instead of computing 2¹⁸, use 18 = 6×3 to write 2¹⁸ = (2⁶)³ = 64³, so the upper bound is 64³+1.
Rewriting 2¹⁸ as (2⁶)³ uses the integer-exponent property (a^m)ⁿ = a^mn — a Grade 8 idea that lets us skip a brute-force computation.
The upper bound 2¹⁸+1 is exactly one more than the perfect cube 64³, because 2¹⁸ itself equals 64³.
▸ Why?
Turning both pieces into 2's shows it: the base 64 is 2⁶, and cubing it lines up eighteen 2's in all, so 64³ = 2¹⁸.
▸ Why?
The base 64 is what you get by multiplying the base 2 by itself six times, and writing that repeated multiplication compactly is exactly what 2⁶ means.
▸ Why?
Cubing 2⁶ means multiplying three copies of it, 2⁶ · 2⁶ · 2⁶, so the 2's inside number 6+6+6.
▸ Why?
Set the three groups of six 2's end to end with none lost or counted twice, and the total is just the parts added: 6+6+6.
▸ Why?
Adding 6 three times is the same as 6 × 3 = 18, since multiplying is adding equal groups over and over.
Find the largest cube
Compare: 64³ ≤ 64³+1 (included) but 65³ ＞ 64³+1 (excluded), so the largest base is n_max = 64.
Picking the largest n satisfying n³ ≤ 64³ + 1 is again testing the inequality on consecutive integers.
6.EE.B.5Identify SubproblemsCount the cubes in range
Subproblem C: count integers n = 7, 8, …, 64. Inclusive count is 64 − 7 + 1 = 58, matching choice (E).
Counting how many whole numbers sit between two values, inclusive, is a Grade 4 multi-step word-problem move.
4.OA.A.3Identify SubproblemsThis AMC 8 problem only needs Grade 8 exponent rules — like (2⁶)³ = 2¹⁸ — you already know!
- Find the smallest cube
- Set the lower index
- Rewrite the upper bound
- Find the largest cube
- Count the cubes in range
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