Difference between revisions of "2013 AIME II Problems/Problem 11"
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+ | ==Problem 11== | ||
+ | Let <math>A = \{1, 2, 3, 4, 5, 6, 7\}</math>, and let <math>N</math> be the number of functions <math>f</math> from set <math>A</math> to set <math>A</math> such that <math>f(f(x))</math> is a constant function. Find the remainder when <math>N</math> is divided by <math>1000</math>. | ||
==Solution== | ==Solution== | ||
Any such function can be constructed by distributing the elements of <math>A</math> on three tiers. | Any such function can be constructed by distributing the elements of <math>A</math> on three tiers. |
Latest revision as of 18:49, 18 August 2021
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
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 .
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 | ||
All AIME Problems and Solutions |
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