AMC 8 · 2018 · #25
Grade 8 number-theoryProblem
How many perfect cubes lie between 28+1 and 218+1, inclusive?
Pick an answer.
AMC 8 2018 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: Count every perfect cube $n^3$ (where $n$ is a positive integer) that lies in the closed interval $[2^8+1,\; 2^{18}+1]$. The endpoints themselves count if they happen to be perfect cubes.
Givens: Lower bound: $2^8 + 1 = 257$; Upper bound: $2^{18} + 1$; Interval is inclusive on both ends; Answer choices: (A) $4$, (B) $9$, (C) $10$, (D) $57$, (E) $58$
Unknowns: The number of integers $n \ge 1$ such that $2^8 + 1 \le n^3 \le 2^{18} + 1$
Understand
Restated: Count every perfect cube $n^3$ (where $n$ is a positive integer) that lies in the closed interval $[2^8+1,\; 2^{18}+1]$. The endpoints themselves count if they happen to be perfect cubes.
Givens: Lower bound: $2^8 + 1 = 257$; Upper bound: $2^{18} + 1$; Interval is inclusive on both ends; Answer choices: (A) $4$, (B) $9$, (C) $10$, (D) $57$, (E) $58$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #6 Guess and Check, #9 Solve an Easier Related Problem
"How many cubes lie in $[L, U]$?" naturally splits (Tool #7) into three subproblems: (a) find the smallest integer $n_{\min}$ with $n_{\min}^3 \ge L$, (b) find the largest integer $n_{\max}$ with $n_{\max}^3 \le U$, (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 $6^3, 7^3$) lands instantly. For (b) the upper bound $2^{18}+1$ looks scary, but Tool #9 (rewrite to an easier form) reveals the trick: $18 = 6 \times 3$, so $2^{18} = (2^6)^3 = 64^3$ — a perfect cube already sitting just below the upper bound.
Execute — Answer: E
6.EE.A.1 Step 1 - Subproblem A — find the smallest cube at or above the lower bound.
- Compute $2^8 + 1 = 256 + 1 = 257$, then test small cubes by guess-and-check.
💡 Evaluating $2^8$ and small whole-number cubes uses whole-number exponent expressions — a Grade 6 skill.
6.EE.B.5 Step 2 - So the smallest perfect cube in the interval is $7^3 = 343$.
- Any cube with base $n \le 6$ falls below $257$ and is excluded.
💡 Choosing $n_{\min} = 7$ is finding the smallest $n$ that makes the inequality $n^3 \ge 257$ true — a Grade 6 "solve an inequality by testing" move.
8.EE.A.1 Step 3 - Subproblem B — find the largest cube at or below the upper bound.
- Directly computing $2^{18}$ is painful, so rewrite using the exponent rule $(a^m)^n = a^{mn}$.
- Since $18 = 6 \times 3$, we have $2^{18} = (2^6)^3 = 64^3$.
- Therefore the upper bound is $64^3 + 1$, which is just $1$ more than the perfect cube $64^3$.
💡 Rewriting $2^{18}$ as $(2^6)^3$ uses the integer-exponent property $(a^m)^n = a^{mn}$ — a Grade 8 idea that lets us skip a brute-force computation.
6.EE.B.5 Step 4 - Compare $64^3$ and $65^3$ to the upper bound.
- $64^3 \le 64^3 + 1$ (included), and $65^3 = (64+1)^3 > 64^3 + 1$ (excluded).
- So the largest valid base is $64$.
💡 Picking the largest $n$ satisfying $n^3 \le 64^3 + 1$ is again testing the inequality on consecutive integers.
4.OA.A.3 Step 5 - Subproblem C — count the integers $n = 7, 8, 9, \dots, 64$.
- The count of consecutive integers from $a$ to $b$ inclusive is $b - a + 1$.
- So the number of perfect cubes is $64 - 7 + 1 = 58$.
- This matches choice $\textbf{(E)}$.
💡 Counting how many whole numbers sit between two values, inclusive, is a Grade 4 multi-step word-problem move.
6.EE.A.1 Subproblem A — find the smallest cube at or above the lower bound. Compute $2^8 6.EE.B.5 So the smallest perfect cube in the interval is $7^3 = 343$. Any cube with base 8.EE.A.1 Subproblem B — find the largest cube at or below the upper bound. Directly compu 6.EE.B.5 Compare $64^3$ and $65^3$ to the upper bound. $64^3 \le 64^3 + 1$ (included), an 4.OA.A.3 Subproblem C — count the integers $n = 7, 8, 9, \dots, 64$. The count of consecu Review
Reasonableness: The smallest cube is $7^3$ and the largest is $64^3$. 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".
Alternative: Tool #3 (Eliminate Possibilities) on the choices: (A) $4$, (B) $9$, (C) $10$ are all far too small — the interval spans from about $257$ to over $260{,}000$, which contains many cubes. That leaves only (D) $57$ and (E) $58$, separated exactly by the inclusive/exclusive endpoint question. Verifying that BOTH endpoints land on or next to cubes ($7^3 = 343$ is inside, and $64^3$ is exactly $1$ below the upper bound) confirms the inclusive count is (E) $58$.
CCSS standards used (min grade 8)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Counting the integers from $7$ to $64$ inclusive using $64 - 7 + 1 = 58$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating $2^8 = 256$, $6^3 = 216$, and $7^3 = 343$ to locate the lower bound and the smallest cube in range.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Choosing the smallest $n$ with $n^3 \ge 257$ and the largest $n$ with $n^3 \le 64^3 + 1$ by testing successive integers.)8.EE.A.1Know and apply the properties of integer exponents (Rewriting $2^{18} = (2^6)^3 = 64^3$ via the exponent rule $(a^m)^n = a^{mn}$, which makes the upper bound a perfect cube.)
⭐ This AMC 8 problem only needs Grade 8 exponent rules — like $(2^6)^3 = 2^{18}$ — you already know!
⭐ This AMC 8 problem only needs Grade 8 exponent rules — like $(2^6)^3 = 2^{18}$ — you already know!
More like this
Same archetype — closest grade level first.
