Difference between revisions of "2004 AMC 10B Problems/Problem 4"
m (→Problem) |
(→Solution 1) |
||
Line 6: | Line 6: | ||
==Solution== | ==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 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>. | 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>. | + | Clearly <math>L=60</math> and the answer is <math>720/60=\boxed{\mathrm{(B)}\ 12}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 21:04, 22 July 2014
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
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, can not 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 | |
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.