Difference between revisions of "2004 AMC 10B Problems/Problem 4"

Revision as of 21:48, 12 June 2009

Problem

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?

$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$

Solution

Solution 1

The product of all six numbers is $6!=720$ . The products of numbers that can be visible are $720/1$ , $720/2$ , ..., $720/6$ . The answer to this problem is their greatest common divisor -- which is $720/L$ , where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$ . Clearly $L=60$ and the answer is $720/60 = \boxed{12}$ .

Solution 2

Clearly, $P$ can not have a prime factor other than $2$ , $3$ and $5$ .

We can not guarantee that the product will be divisible by $5$ , as the number $5$ can end on the bottom.

We can guarantee that the product will be divisible by $3$ (one of $3$ and $6$ will always be visible), but not by $3^2$ .

Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by $2^2$ . This is the most we can guarantee, as when the $4$ is on the bottom side, the two visible even numbers are $2$ and $6$ , and their product is not divisible by $2^3$ .

Hence $P=3\cdot 2^2 = \boxed{12}$ .

Revision as of 12:14, 7 February 2009 (view source) Misof (talk \| contribs) (previous solution was wrong!) ← Older edit		Revision as of 21:48, 12 June 2009 (view source) BillyCollins (talk \| contribs) (→‎See also) Newer edit →
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	== See also ==		== See also ==

−	{{AMC10 box\|year=2004\|ab=B\|num-b=1\|num-a=3}}	+	{{AMC10 box\|year=2004\|ab=B\|num-b=3\|num-a=5}}

2004 AMC 10B (Problems • Answer Key • Resources)
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