2013 AIME II Problems/Problem 11
Problem 11
Let , and let be the number of functions from set to set such that is a constant function. Find the remainder when is divided by .
Solution 1
Any such function can be constructed by distributing the elements of on three tiers.
The bottom tier contains the constant value, for any . (Obviously .)
The middle tier contains elements such that , where .
The top tier contains elements such that equals an element on the middle tier.
There are choices for . Then for a given , there are ways to choose the elements on the middle tier, and then ways to draw arrows down from elements on the top tier to elements on the middle tier.
Thus , giving the answer .
Solution 1 Clarified
Define the three layers as domain , codomain , and codomain . Each one of them is contained in the set . We know that is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value in ways. So now, we choose the values that can be for all those values should satisfy . Let's be that set of values. First things first, we must have to be part of , for the is part of the domain of . Since the values in all satisfy , we have to be a value that can be. Now, for the elements other than :
If we have elements other than that can be part of , we will have ways to choose those values. There will also be ways for each of the elements in other than and those in set (for when function is applied on those values, we already know it would be ). There are elements in other than and those in set . Thus, there should be ways to match the domain to the values of . Summing up all possible values of (), we have
Multiplying that by the original for the choice of , we have
~sml1809
Video Solution
https://youtu.be/aaO7abKG0BQ?si=KLfz6oyzVR0d8D13
~MathProblemSolvingSkills.com
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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