Question

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P?$

$6$

$12$

$24$

$144$

$720$

Answer: B

Concepts:prime factorization divisibility bounding to limit cases

Difficulty rating: 1190

Small Hint:

The product of all six faces is $6! = 720$ ; exactly one face is hidden

Big Hint:

For each prime, find how many copies must remain no matter which face is hidden

Solution:

Since $6! = 720 = 2^4 \cdot 3^2 \cdot 5,$ the visible product uses only the primes $2,$ $3,$ and $5.$

Hiding $4$ leaves the fewest $2$ ’s, namely $2^2.$ Hiding $3$ or $6$ leaves the fewest $3$ ’s, namely one. Hiding $5$ leaves no factor of $5.$

Therefore $P$ is always divisible by $2^2 \cdot 3 = 12,$ but not necessarily by any larger number.

Thus, the correct answer is B.

2004 AMC 10B Problem 4

Answer: B

Difficulty rating: 1190

Small Hint:

Big Hint:

Solution:

Problem 4 in Other Years