Difference between revisions of "2013 AIME II Problems/Problem 11"
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Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>. | Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>. | ||
− | ==Solution | + | ==Solution 1 Clarified== |
+ | |||
+ | Define the three layers as [domain] <math>x</math>, [codomain] <math>f(x)</math>, and codomain <math>f(f(x))</math>. Each one of them is contained in the set <math>A</math>. We know that <math>f(f(x))</math> is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value <math>c</math> in <math>7</math> ways. So now, we choose the values that can be <math>f(x)</math> for all those values should satisfy <math>f(f(x))=c</math>. Let's <math>S</math> be that set of values. First things first, we must have <math>5</math> to be part of <math>S</math>, for the <math>S</math> is part of the domain of <math>x</math>. Since the values in <math>i\in S</math> all satisfy <math>f(i) = 5</math>, we have <math>5</math> to be a value that <math>f(x)</math> can be. Now, for the elements other than <math>5</math>: | ||
+ | |||
+ | If we have <math>k</math> elements other than <math>5</math> than can be part of <math>S</math>, we will have <math>\binom{6}{k}</math> ways to choose those values. There will also be <math>k</math> ways for each of the elements in <math>A</math> other than <math>5</math> and those in set <math>S</math> (for when function <math>f</math> is applied on those values, we already know it would be <math>5</math>). There are <math>6-k</math> elements in <math>A</math> other than <math>5</math> and those in set <math>S</math>. So, there should be <math>6^{6-k}</math> ways to match the domain <math>x</math> to the values of <math>f(x)</math>. So, summing up all possible values of <math>k</math> (<math>[1,6]</math>), we have | ||
+ | |||
+ | <cmath>\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057</cmath> | ||
+ | |||
+ | Multiplying that by the original <math>7</math> for the choice of <math>c</math>, we have <math>7 \cdot 1057 = 7\boxed{399}.</math> | ||
+ | |||
+ | ~sml1809 | ||
==See Also== | ==See Also== |
Revision as of 15:53, 8 January 2023
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
.
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
than 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
. So, there should be
ways to match the domain
to the values of
. So, summing up all possible values of
(
), we have
Multiplying that by the original for the choice of
, we have
~sml1809
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.