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

(New page: In a six sided dice, the only numbers with shared factors are 2,3,6. Since 6 is must either by 2,3 or 6, P must be divisible by 6.)
 
(→‎Solution 3)
 
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In a six sided dice, the only numbers with shared factors are 2,3,6.
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==Problem==
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Since 6 is must either by 2,3 or 6, P must be divisible by 6.
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A standard six-sided die is rolled, and <math>P</math> is the product of the five numbers that are visible. What is the largest number that is certain to divide <math>P</math>?
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<math> \mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720 </math>
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== Solution 1 ==
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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>.
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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>.
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Clearly <math>L=60</math> and the answer is <math>720/60=\boxed{\mathrm{(B)}\ 12}</math>.
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== Solution 2 ==
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Clearly, <math>P</math> cannot have a prime factor other than <math>2</math>, <math>3</math> and <math>5</math>.
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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.
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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>.
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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>.
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== Solution 3 ==
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The product P can be one of the following six numbers excluding the number that is hidden under, so we have:
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\begin{align*}
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2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\
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1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\
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1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\
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1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\
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1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\
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1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5
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\end{align*}
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The largest number that is certain to divide product P is basically GCD of all the above 6 products which is <math>2^2 \cdot 3</math>.
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Hence <math>P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}</math>.
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== See also ==
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{{AMC10 box|year=2004|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 22:42, 16 January 2024

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 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 $720/60=\boxed{\mathrm{(B)}\ 12}$.

Solution 2

Clearly, $P$ cannot 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 $2^3$.

Solution 3

The product P can be one of the following six numbers excluding the number that is hidden under, so we have: \begin{align*} 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\ 1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5 \end{align*}

The largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \cdot 3$.

Hence $P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}$.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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