Competition · AMC preparation · step 4 of 4
AMC 8 · 2018 · #18
Grade 6 number-theoryPick an answer.
AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Listing all divisors of 23,232 by hand is hopeless. Tool #7 (Identify Subproblems) splits the job into two clean pieces: (a) find the prime factorization of 23,232, then (b) turn that factorization into a divisor count. To justify the counting step we lean on Tool #9 (Easier Problem) and Tool #5 (Pattern): try the same procedure on a small number like 12 = 2² · 3, list its 6 divisors, notice that 6 = (2+1)(1+1), and generalize. That's much friendlier than memorizing a formula and keeps the reasoning at an elementary level.
Pull out the factors of 2
23,232 is even, so keep dividing by 2 until it turns odd — that happens six times, leaving 23232 = 2⁶ · 363.
Dividing a multi-digit number by a one-digit divisor over and over is exactly the Grade 5 long-division skill.
5.NBT.B.6Identify SubproblemsFactor the odd part
The odd part 363 = 3 × 121 = 3 × 11² (digit sum 12 shows 3 divides it), so 23232 = 2⁶ · 3¹ · 11².
Recognizing 3 ∣ 363 via the digit-sum rule and spotting 121 = 11² uses the Grade 4 "factor pairs and prime/composite" idea directly.
4.OA.B.4Identify SubproblemsTest the rule on 12
Check on 12 = 2² · 3: it has six divisors, and (2+1)(1+1) = 6 — each prime's power is chosen independently, so the counts multiply.
Solving a tiny version first is the Grade 4 "find all factor pairs" idea — and it makes the multiplicative shortcut obvious.
4.OA.B.4Solve An Easier Related ProblemApply the divisor formula
For 23232 = 2⁶ · 3¹ · 11², the exponents give 7, 2, and 3 independent choices, so the divisor count is (6+1)(1+1)(2+1) = 42.
Reading exponents off a prime factorization p^e is the Grade 6 "whole-number exponents" idea — once read off, the count is plain multiplication.
The number of positive factors of 2⁶ · 3¹ · 11² is exactly the product (6+1)(1+1)(2+1).
▸ Why?
Counting the factors is the same as counting how many ways you can separately choose a power of 2 (from 2⁰ up to 2⁶), a power of 3 (from 3⁰ up to 3¹), and a power of 11 (from 11⁰ up to 11²) and multiply them.
▸ Why?
The factors and these exponent choices line up one for one: every factor is exactly one product 2^a · 3^b · 11^c with 0 ≤ a ≤ 6, 0 ≤ b ≤ 1, 0 ≤ c ≤ 2, and every such product is a factor.
▸ Why?
Because 2⁶ · 3¹ · 11² is the only way to build this number out of primes, any factor can use only the primes 2, 3, and 11, and no more of each than the number itself carries, so the factors are precisely the products 2^a · 3^b · 11^c within those exponent bounds — none left out and none counted twice.
▸ Why?
Once each factor is matched with exactly one exponent choice and each exponent choice with exactly one factor, the two collections have to be the same size, so counting the exponent choices counts the factors.
▸ Why?
There are 7 ways to choose the power of 2, and for each of those the very same 2 × 3 = 6 ways to finish choosing the powers of 3 and 11, so the factors fall into 7 equal groups of 6 and the total is 7 · 2 · 3.
Match 42 to the choices
The count 42 matches answer choice (E).
Compare the result to the listed options — a Grade 4 "factor count" sanity check.
4.OA.B.4Identify SubproblemsThis AMC 8 problem only needs Grade 6 whole-number exponents and a sprinkle of Grade 4 prime-factor know-how that you already have!
- Pull out the factors of 2
- Factor the odd part
- Test the rule on 12
- Apply the divisor formula
- Match 42 to the choices
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