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

Revision as of 12:14, 7 February 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 $P=3\cdot 2^2 = \boxed{12}$ .

@@ Line 1: / Line 1: @@
-In a six sided dice, the only numbers with shared factors are 2,3,6.
+==Problem==
-Since 6 is must either by 2,3 or 6, P must be divisible by 6.
+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?
+<math> \mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720 </math>
+==Solution==
+===Solution 1===
+The product of all six numbers is <math>6!=720</math>. The products of numbers that can be visible are <math>720/1</math>, <math>720/2</math>, ..., <math>720/6</math>.
+The answer to this problem is their greatest common divisor -- which is <math>720/L</math>, where <math>L</math> is the least common multiple of <math>\{1,2,3,4,5,6\}</math>.
+Clearly <math>L=60</math> and the answer is <math>720/60 = \boxed{12}</math>.
+===Solution 2===
+Clearly, <math>P</math> can not have a prime factor other than <math>2</math>, <math>3</math> and <math>5</math>.
+We can not guarantee that the product will be divisible by <math>5</math>, as the number <math>5</math> can end on the bottom.
+We can guarantee that the product will be divisible by <math>3</math> (one of <math>3</math> and <math>6</math> will always be visible), but not by <math>3^2</math>.
+Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by <math>2^2</math>. This is the most we can guarantee, as when the <math>4</math> is on the bottom side, the two visible even numbers are <math>2</math> and <math>6</math>, and their product is not divisible by <math>2^3</math>.
+Hence <math>P=3\cdot 2^2 = \boxed{12}</math>.
+== See also ==
+{{AMC10 box|year=2004|ab=B|num-b=1|num-a=3}}

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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