2013 AIME II Problem 11

Attempt Problem 11 of the 2013 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AIME II solutions, or check the answer key.

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11.

Let A={1,2,3,4,5,6,7},A = \{1, 2, 3, 4, 5, 6, 7\}, and let NN be the number of functions ff from set AA to set AA such that f(f(x))f(f(x)) is a constant function. Find the remainder when NN is divided by 1000.1000.

Answer: 399
Concepts:functioncombinationscasework
Difficulty rating: 2890
Solution:

Say f(f(x))=af(f(x)) = a for all x,x, and let S={x:f(x)=a}.S = \{x : f(x) = a\}. Picking any tS,t \in S, we get a=f(f(t))=f(a),a = f(f(t)) = f(a), so aS.a \in S. Every xx satisfies f(x)Sf(x) \in S (because f(f(x))=af(f(x)) = a), and if xSx \notin S then f(x)a,f(x) \ne a, so ff maps the complement of SS into S{a}.S \setminus \{a\}. Conversely, any ff built this way works.

If S=k,|S| = k, we choose the constant aa in 77 ways, the remaining k1k - 1 elements of SS in (6k1)\binom{6}{k-1} ways, and an image in S{a}S \setminus \{a\} for each of the 7k7 - k other elements in (k1)7k(k-1)^{7-k} ways. Hence N=7k=17(6k1)(k1)7k=7(0+6+240+540+240+30+1)=71057=7399. \begin{aligned} N &= \scriptsize 7\sum_{k=1}^{7} \binom{6}{k-1}(k-1)^{7-k} \\ &\scriptsize = 7\,(0 + 6 + 240 + 540 + 240 + 30 + 1) \\ &= 7 \cdot 1057 = 7399. \end{aligned}

The remainder when NN is divided by 10001000 is 399.399.

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