Difference between revisions of "2010 AMC 12B Problems/Problem 16"
(→See also) |
|||
Line 21: | Line 21: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 10:59, 4 July 2013
Problem 16
Positive integers ,
, and
are randomly and independently selected with replacement from the set
. What is the probability that
is divisible by
?
Solution
The value of is arbitrary other than it is divisible by
, so the set
can be grouped into threes.
Obviously, if is divisible by
(which has probability
) then the sum is divisible by
. In the event that
is not divisible by
(which has probability
, then the sum is divisible by
if
, which is the same as
.
This only occurs when one of the factors or
is equivalent to
and the other is equivalent to
. All four events
,
,
, and
have a probability of
because the set is grouped in threes.
In total the probability is
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.