AMC 10 · 2005 · #15
Grade 8 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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! · 5! · 7! as one prime factorization 2⁸ · 3⁴ · 5² · 7¹. 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, … up to the cap), and the count of whole combinations is just the product of the four separate counts.
Factor each factorial into primes
Break each factorial into prime factors: 3! = 6, 5! = 120, 7! = 5040.
Every whole number is built from prime bricks, so factoring first shows exactly which bricks are available.
4.OA.B.4Identify SubproblemsCombine into one prime factorization
Add the exponents of each shared prime: the 2's give 2¹⁺³⁺⁴, the 3's give 3¹⁺¹⁺², the 5's give 5⁰⁺¹⁺¹, and 7 appears once.
Multiplying powers of the same prime just piles up the exponents, so counting each prime once keeps the bookkeeping clean.
8.EE.A.1Identify SubproblemsSay what makes a divisor a cube
Any divisor is 2^a·3^b·5^c·7^d with a ≤ 8, b ≤ 4, c ≤ 2, d ≤ 1, and it is a cube exactly when all four exponents are multiples of 3.
Cubing triples every exponent, so a cube can only have exponents that come in groups of three.
Cubing triples every exponent, so a cube can only have exponents that come in groups of three.
▸ Why?
An exponent counts how many times a prime is used, so cubing uses each of them three times over.
▸ Why?
Every number has exactly one prime recipe, so each prime's count can be checked on its own.
Count cube-legal exponents per prime
Under each cap list the multiples of 3, including 0: 2 gives 0, 3, 6; 3 gives 0, 3; 5 and 7 give only 0.
You can only take the prime in whole triples, and never more than the number actually contains.
4.OA.B.4Make A Systematic ListMultiply the choices together
The four exponents are chosen independently, so multiply the separate counts: 3 × 2 × 1 × 1.
Independent choices multiply, just like pairing every shirt with every pair of pants.
7.SP.C.8Identify SubproblemsTo 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.
- Factor each factorial into primes
- Combine into one prime factorization
- Say what makes a divisor a cube
- Count cube-legal exponents per prime
- Multiply the choices together