2004 AMC 10B Problem 4

Attempt Problem 4 of the 2004 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

A standard six-sided die is rolled, and PP is the product of the five numbers that are visible. What is the largest number that is certain to divide P?P?
66
1212
2424
144144
720720
Answer: B
Concepts:prime factorizationdivisibilitybounding to limit cases
Small hint:
The product of all six faces is 6!=7206! = 720; exactly one face is hidden
Big hint:
For each prime, find how many copies must remain no matter which face is hidden
Solution:
Since 6!=720=24⋅32⋅5,6! = 720 = 2^4 \cdot 3^2 \cdot 5, the visible product uses only the primes 2,2, 3,3, and 5.5.
Hiding 44 leaves the fewest 22’s, namely 22.2^2. Hiding 33 or 66 leaves the fewest 33’s, namely one. Hiding 55 leaves no factor of 5.5.
Therefore PP is always divisible by 22⋅3=12,2^2 \cdot 3 = 12, but not necessarily by any larger number.
Thus, the correct answer is B.
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