AMC 8 · 2018 · #18
Grade 6 number-theoryProblem
How many positive factors does 23,232 have?
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 positive whole number that divides $23{,}232$ evenly (including $1$ and $23{,}232$ itself), then pick the matching answer choice.
Givens: The target number is $23{,}232$; Answer choices: (A) $9$, (B) $12$, (C) $28$, (D) $36$, (E) $42$; We need the count of positive factors, not the factors themselves
Unknowns: How many positive integers divide $23{,}232$
Understand
Restated: Count every positive whole number that divides $23{,}232$ evenly (including $1$ and $23{,}232$ itself), then pick the matching answer choice.
Givens: The target number is $23{,}232$; Answer choices: (A) $9$, (B) $12$, (C) $28$, (D) $36$, (E) $42$; We need the count of positive factors, not the factors themselves
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #5 Look for a Pattern
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 $12 = 2^2 \cdot 3$, 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.
Execute — Answer: E
5.NBT.B.6 Step 1 - Strip out all factors of $2$ first.
- The number $23{,}232$ is even, so divide by $2$ repeatedly until the result is odd.
- Each successful division contributes one more factor of $2$.
💡 Dividing a multi-digit number by a one-digit divisor over and over is exactly the Grade 5 long-division skill.
4.OA.B.4 Step 2 - Factor the odd part, $363$.
- The digit sum $3+6+3=12$ is divisible by $3$, so $363$ is too.
- Dividing gives $363 \div 3 = 121$, and $121 = 11 \times 11 = 11^2$ is a well-known square.
💡 Recognizing $3 \mid 363$ via the digit-sum rule and spotting $121 = 11^2$ uses the Grade 4 "factor pairs and prime/composite" idea directly.
4.OA.B.4 Step 3 - Before applying any formula, test the counting rule on an easier number to see why it works.
- Take $12 = 2^2 \cdot 3^1$ and list every divisor: $1, 2, 3, 4, 6, 12$ — that's $6$ divisors.
- Notice $(2+1)(1+1) = 3 \cdot 2 = 6$.
- Each divisor of $12$ is built by picking how many $2$'s ($0$, $1$, or $2$ choices) and how many $3$'s ($0$ or $1$ choices), so the count multiplies.
💡 Solving a tiny version first is the Grade 4 "find all factor pairs" idea — and it makes the multiplicative shortcut obvious.
6.EE.A.1 Step 4 - Apply the same idea to $23{,}232 = 2^6 \cdot 3^1 \cdot 11^2$.
- The exponent of $2$ in any divisor can be $0,1,2,3,4,5,$ or $6$ ($7$ choices); the exponent of $3$ can be $0$ or $1$ ($2$ choices); the exponent of $11$ can be $0,1,$ or $2$ ($3$ choices).
- Each independent choice multiplies together.
💡 Reading exponents off a prime factorization $p^e$ is the Grade 6 "whole-number exponents" idea — once read off, the count is plain multiplication.
4.OA.B.4 Step 5 - Match the count $42$ to the answer choices.
- Choice (E) is $42$.
💡 Compare the result to the listed options — a Grade 4 "factor count" sanity check.
5.NBT.B.6 Strip out all factors of $2$ first. The number $23{,}232$ is even, so divide by 4.OA.B.4 Factor the odd part, $363$. The digit sum $3+6+3=12$ is divisible by $3$, so $36 4.OA.B.4 Before applying any formula, test the counting rule on an easier number to see w 6.EE.A.1 Apply the same idea to $23{,}232 = 2^6 \cdot 3^1 \cdot 11^2$. The exponent of $2 4.OA.B.4 Match the count $42$ to the answer choices. Choice (E) is $42$. Review
Reasonableness: Quick sanity check on the prime factorization: $2^6 = 64$, $64 \times 3 = 192$, and $192 \times 121 = 192 \times 121 = 23{,}232$ (since $192 \times 100 = 19{,}200$ and $192 \times 21 = 4{,}032$, summing to $23{,}232$). Also, $42$ is the largest answer choice — that fits a number with three different prime factors and the relatively large exponent $6$ on the prime $2$. A number this size with so much $2$-content should have many divisors, so $42$ is far more reasonable than the smaller choices like $9$ or $12$.
Alternative: Tool #2 (Systematic List) gives a brute-force backup: once the factorization $2^6 \cdot 3 \cdot 11^2$ is in hand, list divisors in an organized grid by powers of $11$. The $11^0$ row pairs every divisor of $2^6 \cdot 3$ ($14$ of them) with $1$; the $11^1$ row repeats with a factor of $11$; the $11^2$ row repeats with $121$. Total $= 14 + 14 + 14 = 42$ — the same answer, reached without quoting the multiplicative shortcut as a formula.
CCSS standards used (min grade 6)
5.NBT.B.6Find whole-number quotients with up to four-digit dividends and two-digit divisors (Performing the chain of long divisions $23232 \div 2 \div 2 \div \cdots$ and $363 \div 3$ to strip primes off $23{,}232$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that $2$, $3$, and $11$ are primes, using the digit-sum test for divisibility by $3$, identifying $121 = 11^2$, and trying the rule on the smaller case $12 = 2^2 \cdot 3$ to see why the divisor count multiplies.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Reading exponents from the prime factorization $23{,}232 = 2^6 \cdot 3^1 \cdot 11^2$ and turning each exponent $e$ into a count of $e+1$ choices, then multiplying $(6+1)(1+1)(2+1) = 42$.)
⭐ This AMC 8 problem only needs Grade 6 whole-number exponents and a sprinkle of Grade 4 prime-factor know-how that you already have!
⭐ This AMC 8 problem only needs Grade 6 whole-number exponents and a sprinkle of Grade 4 prime-factor know-how that you already have!
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