Mock AIME 3 Pre 2005 Problems/Problem 13
Contents
[hide]Problem
Let
denote the value of the sum
Determine the remainder obtained when is divided by
.
Solution 1
Let . Let
. Then note that
, so taking the derivative and multiplying by
gives
. Taking the derivative and multiplying by
again gives
. Now note that
. Then we get
, so
, so
.
Solution 2
Let the wanted sum be .
We will simplify the expression into:
.
A counting argument will be provided to compute this.
Consider people, who will be separated into group
, group
, and group
. Furthermore, one person in group
will be cool, and one person will be smart. (They may be the same people).
Consider only putting people into group
. There are
ways this can be done. For the remaining
people, there are one of two groups they can be in, namely group
and group
. This means that there are
ways this can be done. There are
ways to determine who is cool, and
ways to determine who is smart. This is
. As
ranges from
to
, we will get all such scenarios. This means that the number of ways that this can be done is also
.
Another way to count this is two split it up into two cases.
Case 1: person is both cool and smart.
There are
ways to choose this person. The remaining
people have a choice of one of
groups, making
ways.
Case 2: person is cool, and another person in smart.
There are
ways to choose who is cool, and
ways to choose who is smart. The remaining
people have a choice of one of
groups, making
.
Thus, we have:
Thus, the answer is
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |