2017 AIME I Problems/Problem 7
Contents
Problem 7
For nonnegative integers and
with
, let
. Let
denote the sum of all
, where
and
are nonnegative integers with
. Find the remainder when
is divided by
.
Solution 1
Let , and note that
. The problem thus asks for the sum
over all
such that
. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately, which is equal to
. Therefore, the answer is
.
-rocketscience
Solution 2
Treating as
, this problem asks for
But
can be computed through the following combinatorial argument. Choosing
elements from a set of size
is the same as splitting the set into two sets of size
and choosing
elements from one,
from the other where
. The number of ways to perform such a procedure is simply
. Therefore, the requested sum is
As such, our answer is
.
- Awsomness2000
Solution 3 (Major Major Bash)
Case 1: .
Subcase 1:
Subcase 2:
Subcase 3:
Case 2:
By just switching and
in all of the above cases, we will get all of the cases such that
is true. Therefore, this case is also
Case 3:
Solution 4
We begin as in solution 1 to rewrite the sum as over all
such that
.
Consider the polynomial
.
We can see the sum we wish to compute is just the coefficient of the
term. However
. Therefore, the coefficient of the
term is just
so the answer is
.
- mathymath
Solution 5
Let . Then
, and
. The problem thus asks for
Suppose we have
red balls,
green balls, and
blue balls lined up in a row, and we want to choose
balls from this set of
balls by considering each color separately. Over all possible selections of
balls from this set, there are always a nonnegative number of balls in each color group. The answer is
.
Solution 6
Since , we can rewrite
as
. Consider the number of ways to choose a committee of 6 people from a group of 6 democrats, 6 republicans, and 6 independents. We can first pick
democrats, then pick
republicans, provided that
. Then we can pick the remaining
people from the independents. But this is just
, so the sum of all
is equal to the number of ways to choose this committee.
On the other hand, we can simply pick any 6 people from the
total politicians in the group. Clearly, there are
ways to do this. So the desired quantity is equal to
. We can then compute (routinely) the last 3 digits of
as
.
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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