AMC 10 · 2004 · #4

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisibility-rulesprime-factorizationfactors extremal-constructionsystematic-enumeration ↑ Prerequisites: factors
📏 Medium solution 💡 2 insights
Problem
A standard die shows the numbers 1, 2, 3, 4, 5, 6 on its six faces. It is rolled, so exactly one face rests on the bottom and is hidden, and P is the product of the five visible numbers. Find the largest number that is certain to divide P, no matter which face is hidden.

Pick an answer.

(A)
6
(B)
12
(C)
24
(D)
144
(E)
720

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

"Certain to divide" is a guarantee: the divisor has to survive the worst possible roll, so this is an Extreme Principle (#14) question — for each prime factor, find the roll that leaves the fewest of them. Make a Systematic List (#2) turns the vague "any hidden face" into six concrete products so the worst case is visible, and Eliminate Possibilities (#3) matches the surviving number to a choice while ruling out the larger traps that only work sometimes.

1STEP 1

Fix the total product

All six faces multiply to a fixed total of 720, so hiding one face leaves P equal to that total divided by the hidden number.

1·2·3·4·5·6=720, P=720/(hidden face)
2STEP 2

Turn "certain" into worst case

"Certain" means it must divide P on every roll, so factor 720=2⁴·3²·5 and hunt for the fewest copies of each prime.

720=2⁴·3²·5
3STEP 3

List the six possible products

Write P=720÷h for each hidden face h and factor it, so the worst case for every prime shows up at a glance.

h=1&:720=2⁴·3²·5 ; h=2&:360=2³·3²·5 ; h=3&:240=2⁴·3·5 ; h=4&:180=2²·3²·5 ; h=5&:144=2⁴·3² ; h=6&:120=2³·3·5
4STEP 4

Take the smallest power of each prime

Take the gcd: fewest 2s is 2² (hide 4), fewest 3s is 3¹ (hide 3 or 6), and the 5 disappears (hide 5), leaving 2²·3=12.

gcd=2^min(4,3,4,2,4,3)·3^min(2,2,1,2,2,1)·5^min(1,1,1,1,0,1)=2²·3¹·5⁰=12
5STEP 5

Match to a choice and rule out the traps

12 divides all six products; 24 fails on 180, 144 and 720 lean on a 5 that can vanish, and 6 is not the largest — choice (B).

12 ∣ P for all h, 24 ∤ 180, 5 ∤ 144 → (B)
Answer
12
Check 12 against the extremes of the list: 180 (its smallest value) gives 180÷12=15, and 144 gives 144÷12=12 — both whole, so 12 really does divide every case. Pushing higher breaks: 24 fails on 180 (only two 2s available) and anything with a factor of 5 fails on 144. So 12 is exactly the ceiling, which lands on (B) and rules out the larger (C), (D), (E) while beating (A).
💡Key takeaway

"Certain to divide" means it has to survive the worst roll, so for each prime keep only as many as the poorest case leaves — here that is 2·2·3=12.

  • Fix the total product
  • Turn "certain" into worst case
  • List the six possible products
  • Take the smallest power of each prime
  • Match to a choice and rule out the traps

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