AMC 10 · 2005 · #15
Grade 8 arithmeticHow many positive cubes divide 3!⋅5!⋅7!?
Pick an answer.
AMC 10 2005 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 how many positive perfect cubes are divisors of the number $3! \cdot 5! \cdot 7!$.
Givens: The number is the product of three factorials: $3! \cdot 5! \cdot 7!$; A perfect cube is a number of the form $n^3$ for a positive integer $n$; We only count cubes that divide the product evenly, with no remainder; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The prime factorization of $3! \cdot 5! \cdot 7!$; How many divisors of that number are perfect cubes
Understand
Restated: Count how many positive perfect cubes are divisors of the number $3! \cdot 5! \cdot 7!$.
Givens: The number is the product of three factorials: $3! \cdot 5! \cdot 7!$; A perfect cube is a number of the form $n^3$ for a positive integer $n$; We only count cubes that divide the product evenly, with no remainder; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #5 Look for a Pattern
A cube divisor is decided one prime at a time, so Tool #7 splits the hard question into four easy ones: for each prime $2, 3, 5, 7$, how many cube-legal exponents fit under the cap set by the big number? First rewrite $3! \cdot 5! \cdot 7!$ as one prime factorization $2^8 \cdot 3^4 \cdot 5^2 \cdot 7^1$. Tool #5 supplies the key pattern: a divisor is a cube exactly when each exponent is a multiple of $3$. Tool #2 then lists the legal exponents for each prime ($0, 3, 6, \dots$ up to the cap), and the count of whole combinations is just the product of the four separate counts.
Execute — Answer: E
4.OA.B.4 Step 1 Factor each factorial into primes
- Write out each factorial and break it into prime factors.
- $3! = 6$, $5! = 120$, and $7! = 5040$.
- Splitting each into primes gives the pieces we can recombine.
💡 Every whole number is built from prime bricks, so factoring first shows exactly which bricks are available.
8.EE.A.1 Step 2 Combine into one prime factorization
- Multiply the three results by adding the exponents of each shared prime.
- The $2$'s give $2^{1+3+4}$, the $3$'s give $3^{1+1+2}$, the $5$'s give $5^{0+1+1}$, and there is a single $7$.
💡 Multiplying powers of the same prime just piles up the exponents, so counting each prime once keeps the bookkeeping clean.
6.EE.A.1 Step 3 Say what makes a divisor a cube
- Any divisor looks like $2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d}$, where each exponent is at most the exponent in the big number: $a \le 8$, $b \le 4$, $c \le 2$, $d \le 1$.
- Such a divisor is a perfect cube exactly when every exponent is a multiple of $3$.
💡 Cubing triples every exponent, so a cube can only have exponents that come in groups of three.
4.OA.B.4 Step 4 Count cube-legal exponents per prime
- For each prime, list the multiples of $3$ (including $0$) that stay under its cap.
- For $2$ ($\le 8$): $0, 3, 6$ — three choices.
- For $3$ ($\le 4$): $0, 3$ — two choices.
- For $5$ ($\le 2$): only $0$ — one choice.
- For $7$ ($\le 1$): only $0$ — one choice.
💡 You can only take the prime in whole triples, and never more than the number actually contains.
7.SP.C.8 Step 5 Multiply the choices together
The four exponents are chosen independently, so the total number of cube divisors is the product of the separate counts: $3 \times 2 \times 1 \times 1$.
💡 Independent choices multiply, just like pairing every shirt with every pair of pants.
4.OA.B.4 Write out each factorial and break it into prime factors. $3! = 6$, $5! = 120$, 8.EE.A.1 Multiply the three results by adding the exponents of each shared prime. The $2$ 6.EE.A.1 Any divisor looks like $2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d}$, where each e 4.OA.B.4 For each prime, list the multiples of $3$ (including $0$) that stay under its ca 7.SP.C.8 The four exponents are chosen independently, so the total number of cube divisor Review
Reasonableness: The six cubes can be listed explicitly to confirm the count: taking $a \in \{0,3,6\}$ and $b \in \{0,3\}$ (with $c = d = 0$) gives $1,\ 2^3,\ 2^6,\ 3^3,\ 2^3 3^3,\ 2^6 3^3$ — exactly six perfect cubes, each dividing $2^8 \cdot 3^4 \cdot 5^2 \cdot 7$. Primes $5$ and $7$ contribute nothing extra because their exponents ($2$ and $1$) cannot reach $3$, which correctly rules out any cube using $5$ or $7$. The count $6$ matches choice (E).
Alternative: Instead of counting, list every cube divisor directly (Tool #2). The only primes with an exponent of at least $3$ are $2$ (up to $2^6$) and $3$ (up to $3^3$), so the cube divisors are all products of $\{1, 2^3, 2^6\}$ with $\{1, 3^3\}$: that is $1, 8, 64, 27, 216, 1728$ — six numbers, agreeing with the answer.
CCSS standards used (min grade 8)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Breaking each factorial into prime factors and listing the multiples of $3$ available for each prime's exponent.)8.EE.A.1Know and apply the properties of integer exponents (Adding exponents of the same prime when multiplying the three factorials into a single factorization $2^8 \cdot 3^4 \cdot 5^2 \cdot 7$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Writing a general divisor as $2^a 3^b 5^c 7^d$ and stating the perfect-cube condition that every exponent be a multiple of $3$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Applying the multiplication counting principle to combine the independent per-prime exponent choices into the total $3 \times 2 \times 1 \times 1$.)
⭐ To count cube divisors, prime-factorize the number, then for each prime count how many multiples of $3$ (including $0$) fit under its exponent — and multiply those counts.
⭐ To count cube divisors, prime-factorize the number, then for each prime count how many multiples of $3$ (including $0$) fit under its exponent — and multiply those counts.
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