AMC10 2004 B

AMC10 2004 B · Q4

AMC10 2004 B · Q4. It mainly tests Divisibility & factors, Primes & prime factorization.

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 $P$?

掷一个标准的六面骰子，$P$ 是可见的五个数字的乘积。什么是最保证能整除 $P$ 的最大数？

(A) 6 6

(B) 12 12

(D) 144 144

(E) 720 720

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since $6! = 720 = 24 \cdot 3^2 \cdot 5$, the prime factors of $P$ can consist of at most 2’s, 3’s, and 5’s. The least possible number of 2’s is two, which occurs when 4 is not visible. The least possible number of 3’s is one, which occurs when either 3 or 6 is not visible, and the least number of 5’s is zero, when 5 is not visible. Thus $P$ must be divisible by $2^2 \cdot 3 = 12$, but not necessarily by any larger number.

因为 $6! = 720 = 24 \cdot 3^2 \cdot 5$，$P$ 的质因数最多包含 2、3 和 5。最少的 2 的个数是两个（当 4 不可见时）。最少的 3 的个数是一个（当 3 或 6 不可见时），最少的 5 的个数是零（当 5 不可见时）。因此 $P$ 一定能被 $2^2 \cdot 3 = 12$ 整除，但不一定被更大的数整除。

Topics