AMC 10 · 2023 · #23

Grade 11 number-theorycounting
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
prime-factorizationdivisor-counttriangular-numberssystematic-enumeration identify-subproblemsconvert-to-algebrasystematic-enumeration ↑ Prerequisites: prime-factorizationdivisor-count
📏 Long solution 💡 5 insights
Problem
Several ordinary six-sided dice are rolled at once and the numbers showing are multiplied together. Different rolls can give the same product, so what matters is how many different values the product can take. That number is exactly 936. Find how many dice there are.

Pick an answer.

(A)
11
(B)
6
(C)
8
(D)
10
(E)
9
How to solve
Strategy Organize Information in More Ways

Listing products directly is hopeless, because the products overlap in messy ways and there are hundreds of them. The escape is to stop storing a product as a number and start storing it as three counts. Every face is built out of the primes 2, 3, and 5 only, so every product is 2^a 3^b 5^c, and unique factorization means the triple (a,b,c) is a perfect stand-in for the product. Now the trap has to be respected: the reachable triples are not the whole rectangular box of exponents, because each die can only do one job, and a die spent on a 5 is a die that is not pushing the power of 2 up. The fix is to organize the count in the right order — fix b and c first, since those two are literally headcounts of dice, and then the leftover dice make a sweep a gap-free run whose length is easy to write down. Summing those run lengths gives a cubic in n, and one factorization finishes it.

1STEP 1

Store it as three exponents

Store the product as three exponents.

1 = 2⁰ 3⁰ 5⁰, 2 = 2¹, 3 = 3¹, 4 = 2², 5 = 5¹, 6 = 2¹ · 3¹ → product = 2^a 3^b 5^c
2STEP 2

Why the obvious count fails

A naive product double counts.

n = 1: (2 · 1 + 1)(1+1)² = 12 ≠ 6
3STEP 3

Two exponents read straight off

Two are read straight off the roll.

c = (how many dice show 5), b = (how many dice show 3 or 6), b + c ≤ n, k = n - b - c
4STEP 4

What the last exponent sweeps

The last one sweeps a gap-free run.

a_max = b + 2k = b + 2(n - b - c) = 2n - b - 2c, (number of values of a) = 2n + 1 - b - 2c
5STEP 5

Add the runs over every case

Add the run lengths over every case.

N(n) = Σ_b=0ⁿ Σ_c=0ⁿ-b (2n + 1 - b - 2c) = Σ_b=0ⁿ (n - b + 1)(n + 1) = (n+1) · (n+1)(n+2)/2 = (n+1)² (n+2)/2
6STEP 6

Set the formula equal to the target

Set the formula equal to the target.

(n+1)² (n+2)/2 = 936 → (n+1)² (n+2) = 1872
7STEP 7

Factor and read off

Factoring gives 11.

1872 = 2⁴ · 3² · 13 = 12² · 13 = (n+1)² (n+2) → n + 1 = 12, n + 2 = 13, n = 11
Answer
11
The formula can be tested against cases small enough to check by hand. For n = 1 it gives (2² · 3)/2 = 6, and one die really does give the six distinct products 1 through 6. For n = 2 it gives (3² · 4)/2 = 18; two dice give C(7, 2) = 21 unordered pairs, and exactly three collisions occur, namely 1 · 4 = 2 · 2, 1 · 6 = 2 · 3, and 2 · 6 = 3 · 4, leaving 21 - 3 = 18. Running the formula over every listed choice gives n = 6 → 196, n = 8 → 405, n = 9 → 550, n = 10 → 726, and n = 11 → 936. Only n = 11 lands on the target, and since N(n) strictly increases no other n could. A rough size check agrees as well: N(n) behaves like n³/2, so n³ ≈ 1872 and n ≈ 12.3 — slightly high, exactly as expected since the true formula uses n+1 and n+2 in place of n, which drags the answer down to 11. Finally, the naive box count for n = 11 would have been 23 · 12 · 12 = 3312, more than three times 936, which confirms that the reachable triples really are only a slice of the box.
💡Key takeaway

Every dice product is really just an address (a,b,c) saying how many 2s, 3s, and 5s it holds, and the count that matters is the number of addresses the dice can actually reach — not the number of addresses that merely fit inside the box.

  • Store a product as three exponents
  • See why the obvious count fails
  • Read b and c straight off the roll
  • The power of 2 sweeps a gap-free run
  • Add the runs over every case
  • Set the formula equal to 936
  • Factor 1872 and read off n

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