2017 AIME I Problems/Problem 7
Contents
- 1 Problem 7
- 2 Major Note
- 3 Solution 1 (Committee Forming)
- 4 Solution 1 but different (Committee Forming)
- 5 Solution 2 (Committee Forming but slightly more bashy)
- 6 Solution 3 (Major Major Bash)
- 7 Solution 4
- 8 Solution 5 (Committee Forming but different)
- 9 Solution 5 but different (Committee Forming)
- 10 Solution 6(NICE Journal)
- 11 Remark
- 12 See Also
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 .
Major Note
Most solutions use committee forming (except for the bash solution). To understand more about the techniques used, visit the committee forming page for more information.
Solution 1 (Committee Forming)
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 1 but different (Committee Forming)
Alternatively, one can note that we can consider groups where is constant, say . Fix any value of . Then the sum of all of the values of such that is which by Vandermonde's is . Remember, that expression is the resulting sum for a fixed . So, for , we want . This is (by Vandermonde's or committee forming) ~ firebolt360
Note
Now just a quick explanation for people who don't fully understand Vandermonde's. Take the first part, . Consider different groups, and both of size people. We wish to chose peoples from and people from . In total, we chose people. We can then draw a bijection towards choosing people from , which has size . So, it is . Similarly, for , we see that . Now the total is , and the sum is . So, we get . See committee forming for more information ~ firebolt360
Solution 2 (Committee Forming but slightly more bashy)
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 (Committee Forming but different)
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 5 but different (Committee Forming)
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 .
Solution 6(NICE Journal)
Note that . So we have . If we think about this this is essentially choosing a group of people from people, a group of people from people, and a group of from another group of people. This is nothing but choosing people from a group of people. This is nothing but . ~coolmath_2018
Remark
This problem is an example of the generalization of Vandermonde's theorem, which states that for nonnegative and , we have ~eibc
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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