Difference between revisions of "2004 AMC 10B Problems/Problem 4"
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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>. | 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>. | ||
| + | |||
| + | == 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 <math>2^2 \cdot 3</math>. | ||
Hence <math>P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}</math>. | Hence <math>P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}</math>. | ||
Latest revision as of 22:42, 16 January 2024
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 
.
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 
.
Hence 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
