2004 AMC 10B Problems/Problem 4
Contents
Problem
A standard six-sided die is rolled, and is the product of the five numbers that are visible. What is the largest number that is certain to divide ?
Solution 1
The product of all six numbers is . The products of numbers that can be visible are , , ..., . The answer to this problem is their greatest common divisor -- which is , where is the least common multiple of . Clearly and the answer is .
Solution 2
Clearly, cannot have a prime factor other than , and .
We can not guarantee that the product will be divisible by , as the number can end on the bottom.
We can guarantee that the product will be divisible by (one of and will always be visible), but not by .
Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by . This is the most we can guarantee, as when the is on the bottom side, the two visible even numbers are and , and their product is not divisible by .
Hence .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AMC 10 Problems and Solutions |
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