Difference between revisions of "2013 AIME II Problems/Problem 11"

Revision as of 13:07, 28 May 2023

Problem 11

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ .

Solution 1

Any such function can be constructed by distributing the elements of $A$ on three tiers.

The bottom tier contains the constant value, $c=f(f(x))$ for any $x$ . (Obviously $f(c)=c$ .)

The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$ , where $1\le k\le 6$ .

The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier.

There are $7$ choices for $c$ . Then for a given $k$ , there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier.

Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$ , giving the answer $\boxed{399}$ .

Solution 1 Clarified

Define the three layers as domain $x$ , codomain $f(x)$ , and codomain $f(f(x))$ . Each one of them is contained in the set $A$ . We know that $f(f(x))$ is a constant function, or in other words, can only take on one value. So, we can start off by choosing that value $c$ in $7$ ways. So now, we choose the values that can be $f(x)$ for all those values should satisfy $f(f(x))=c$ . Let's $S$ be that set of values. First things first, we must have $5$ to be part of $S$ , for the $S$ is part of the domain of $x$ . Since the values in $i\in S$ all satisfy $f(i) = 5$ , we have $5$ to be a value that $f(x)$ can be. Now, for the elements other than $5$ :

If we have $k$ elements other than $5$ that can be part of $S$ , we will have $\binom{6}{k}$ ways to choose those values. There will also be $k$ ways for each of the elements in $A$ other than $5$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $5$ ). There are $6-k$ elements in $A$ other than $5$ and those in set $S$ . So, there should be $6^{6-k}$ ways to match the domain $x$ to the values of $f(x)$ . So, summing up all possible values of $k$ ( $[1,6]$ ), we have

$\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057$

Multiplying that by the original $7$ for the choice of $c$ , we have $7 \cdot 1057 = 7\boxed{399}.$

~sml1809

Solution 2

Let the domain of $f(x)$ be set $A$ and the image of $f(x)$ be $B$ . Essentially, $f(A) = B$ and $f(B) = c$ , where $c$ is the constant. We now split into cases regarding the number of terms $n$ in $B$ :

$n=1$ : There are $\binom{7}{1}$ ways to choose a term for $B$ . All terms in $A$ map to that 1 term.

$n=2$ : There are $\binom{7}{2}$ ways to choose 2 terms for $B$ . Now, since $f(B) = f(\{1, 2\}) = c$ , $A$ now becomes $\{\{1, 2\}, 3, 4, 5, 6, 7\}$

@@ Line 29: / Line 29: @@
 ==Solution 2==
-Let the domain of <math>f(x)</math> be set <math>A</math> and the image of <math>f(x)</math> be <math>B</math>. Essentially, <math>f(A) = B</math> and <math>f(B) = c</math>, where <math>c</math> is the constant.
+Let the domain of <math>f(x)</math> be set <math>A</math> and the image of <math>f(x)</math> be <math>B</math>. Essentially, <math>f(A) = B</math> and <math>f(B) = c</math>, where <math>c</math> is the constant. We now split into cases regarding the number of terms <math>n</math> in <math>B</math>:
+<math>n=1</math>: There are <math>\binom{7}{1}</math> ways to choose a term for <math>B</math>. All terms in <math>A</math> map to that 1 term.
+<math>n=2</math>: There are <math>\binom{7}{2}</math> ways to choose 2 terms for <math>B</math>. Now, since <math>f(B) = f(\{1, 2\}) = c</math>, <math>A</math> now becomes <math>\{\{1, 2\}, 3, 4, 5, 6, 7\}</math>
 ==See Also==

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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