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

Latest revision as of 22:20, 17 October 2024

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 $S$ be that set of values. First things first, we must have $c$ to be part of $S$ , for the $S$ is part of the domain of $x$ . Since the values in $i\in S$ all satisfy $f(i) = c$ , we have $c$ 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 $c$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $c$ ). There are $6-k$ elements in $A$ other than $c$ and those in set $S$ . Thus, there should be $k^{6-k}$ ways to match the domain $x$ to the values of $f(x)$ . Summing up all possible values of $k$ ( $[1,6]$ ), we have

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

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

Solution 2

It is clear that we must have one fixed point (that is, $f(x)=x$ ). WLOG, let $x=1$ be a fixed point, so $f(1)=1$ .

Now, let's do casework on how many of the inputs $2, 3, 4, 5, 6 ,7$ leads to $1$ . Generally, if some values in that aforementioned list leads to $1$ , then running it in the function again will yield $1$ . All other values must be the the values that leads to $1$ .

For example:

$\textbf{Case 1:}$ All $6$ of $2, 3, 4, 5, 6, 7$ lead to $1$ . In this case, there is only $1$ way.

$\textbf{Case 2:}$ $5$ of $6$ of $2, 3, 4, 5, 6, 7$ lead to $1$ . In this case, we choose $5$ of the $6$ to lead to $1$ : $6\choose5$ .

Then, the other value that does not lead to $1$ should be one of the values that do: $5$ ways.

$\binom{6}{5}\cdot5$

$\textbf{Case 3:}$ $4$ of $6$ lead to $1$ . Choose which lead to $1$ : $6\choose4$ then the other values: $4^2$ ways

$\binom{6}{4}\cdot4^2$

$\textbf{Case 4:}$ $3$ of $6$ lead to $1$ . $\binom{6}{3}\cdot3^3$

$\textbf{Case 5:}$ $2$ of $6$ lead to $1$ . $\binom{6}{2}\cdot2^4$

$\textbf{Case 6:}$ $1$ of $6$ lead to $1$ . $\binom{6}{1}\cdot1^5$

Adding up all the cases, we have $1057$ cases. But don't forget to account for the WLOG and multiply by $7$ , yielding us the final answer of $7\boxed{399}.$

~xHypotenuse

Video Solution

https://youtu.be/aaO7abKG0BQ?si=KLfz6oyzVR0d8D13

~MathProblemSolvingSkills.com

@@ Line 2: / Line 2: @@
 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 1==
-Let the range be <math>R=\{f(x)|x \in A\}</math>
+Any such function can be constructed by distributing the elements of <math>A</math> on three tiers.
-Let the constant function be <math>f(f(x))=c</math>, clearly, <math>c \in R</math>
+The bottom tier contains the constant value, <math>c=f(f(x))</math> for any <math>x</math>. (Obviously <math>f(c)=c</math>.)
-For every <math>a \in R</math>, since <math>f(f(x))=c</math>, we must have <math>f(a)=c</math>.
+The middle tier contains <math>k</math> elements <math>x\ne c</math> such that <math>f(x)=c</math>, where <math>1\le k\le 6</math>.
-Now we enumerate through possible sizes of <math>R</math>. <math>|R|</math> cannot be more than 4 since all the numbers in <math>R</math> must map to <math>c</math> and any number other than <math>c</math> in <math>R</math> must have at least one number mapped to it.
+The top tier contains <math>6-k</math> elements such that <math>f(x)</math> equals an element on the middle tier.
-. <math>|R|=1</math>
+There are <math>7</math> choices for <math>c</math>. Then for a given <math>k</math>, there are <math>\tbinom6k</math> ways to choose the elements on the middle tier, and then <math>k^{6-k}</math> ways to draw arrows down from elements on the top tier to elements on the middle tier.
-We have <math>\binom{1}{7}=7</math> choices for <math>R</math>, <math>\binom{1}{1}=1</math> choices for <math>c</math>, and only 1 way to map the 6 numbers that are not in <math>R</math> (remaining numbers) since the range <math>R</math> is size 1. In total we have <math>7\cdot 1\cdot 1=7</math> choices.
+Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>.
-. <math>|R|=2</math>
+==Solution 1 Clarified==
-We have <math>\binom{2}{7}=21</math> choices for <math>R</math>, <math>\binom{1}{2}=2</math> choices for <math>c</math>.
+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 <math>S</math> be that set of values. First things first, we must have <math>c</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) = c</math>, we have <math>c</math> to be a value that <math>f(x)</math> can be. Now, for the elements other than <math>5</math>:
-Assigning the 5 numbers to the two elements in <math>R</math> yields <math>2^5=32</math> choices. However, one of these choices is to assign all 5 remaining numbers to <math>c</math>, resulting in <math>R</math> being only of size 1. Exclude that and we have <math>32-1=31</math> choices.
+If we have <math>k</math> elements other than <math>5</math> that 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>c</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>c</math>). There are <math>6-k</math> elements in <math>A</math> other than <math>c</math> and those in set <math>S</math>. Thus, there should be <math>k^{6-k}</math> ways to match the domain <math>x</math> to the values of <math>f(x)</math>. Summing up all possible values of <math>k</math> (<math>[1,6]</math>), we have
-In total we have <math>21*2*31=1302</math> choices.
+<cmath>\sum_{k=1}^6 \binom{6}{k} k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1 = 1057</cmath>
-. <math>|R|=3</math>
+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>
-We have <math>\binom{3}{7}=35</math> choices for <math>R</math>, <math>\binom{1}{3}=3</math> choices for <math>c</math>.
-Notice that assigning the remaining numbers is equivalent to distributing 4 numbers into three groups where two of the three groups must receive at least 1 number (since we already have all numbers in <math>R</math> mapped to <math>c</math>, the restriction is not necessary for the <math>c</math> group). Ignoring the restrictions we have <math>3^4</math> ways. Now minus the two cases where one of the two restricted group is left empty. There are 2 ways to choose the left-out group, and <math>2^4</math> ways to distribute numbers for each choice. Finally, by inclusion-exclusion principle, we need to add back the case in which both restricted groups are left empty, which has only <math>1^4=1</math> occurrence. In all, we have <math>3^4-2*2^4+1^4=50</math> ways to assign the remaining numbers.
+==Solution 2==
-In total, we have <math>35*3*50=5250</math> choices for this case.
+It is clear that we must have one fixed point (that is, <math>f(x)=x</math>). WLOG, let <math>x=1</math> be a fixed point, so <math>f(1)=1</math>.
-. <math>|R|=4</math>
+Now, let's do casework on how many of the inputs <math>2, 3, 4, 5, 6 ,7</math> leads to <math>1</math>. Generally, if some values in that aforementioned list leads to <math>1</math>, then running it in the function again will yield <math>1</math>. All other values must be the the values that leads to <math>1</math>.
-We have <math>\binom{4}{7}=35</math> choices for <math>R</math>, <math>\binom{1}{4}=4</math> choices for <math>c</math>.
-Now we have 3 "groups" to fill and only 3 remaining numbers. Thus each group must receive exactly 1 number. However, we still have <math>3!=6</math> ways to permute the mappings.
+For example:
-Thus, we have <math>35*4*6=840</math> choices in total.
-To summarize, we have <math>7+1302+5250+840=7399</math> different mappings possible. So the answer is <math>\boxed{399}</math>.
+<math>\textbf{Case 1:}</math> All <math>6</math> of <math>2, 3, 4, 5, 6, 7</math> lead to <math>1</math>.
+In this case, there is only <math>1</math> way.
+<math>\textbf{Case 2:}</math> <math>5</math> of <math>6</math> of <math>2, 3, 4, 5, 6, 7</math> lead to <math>1</math>.
+In this case, we choose <math>5</math> of the <math>6</math> to lead to <math>1</math>: <math>6\choose5</math>.
+Then, the other value that does not lead to <math>1</math> should be one of the values that do: <math>5</math> ways.
+<math>\binom{6}{5}\cdot5</math>
+<math>\textbf{Case 3:}</math> <math>4</math> of <math>6</math> lead to <math>1</math>.
+Choose which lead to <math>1</math>: <math>6\choose4</math> then the other values: <math>4^2</math> ways
+<math>\binom{6}{4}\cdot4^2</math>
+<math>\textbf{Case 4:}</math> <math>3</math> of <math>6</math> lead to <math>1</math>.
+<math>\binom{6}{3}\cdot3^3</math>
+<math>\textbf{Case 5:}</math> <math>2</math> of <math>6</math> lead to <math>1</math>.
+<math>\binom{6}{2}\cdot2^4</math>
+<math>\textbf{Case 6:}</math> <math>1</math> of <math>6</math> lead to <math>1</math>.
+<math>\binom{6}{1}\cdot1^5</math>
+Adding up all the cases, we have <math>1057</math> cases. But don't forget to account for the WLOG and multiply by <math>7</math>, yielding us the final answer of <math>7\boxed{399}.</math>
+~xHypotenuse
+==Video Solution==
+https://youtu.be/aaO7abKG0BQ?si=KLfz6oyzVR0d8D13
+~MathProblemSolvingSkills.com
-{{solution}}
 ==See Also==
 {{AIME box|year=2013|n=II|num-b=10|num-a=12}}
+{{MAA Notice}}
 [[Category:Intermediate Combinatorics Problems]]

2013 AIME II (Problems • Answer Key • Resources)
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