2023 AMC 12A Problem 24

Attempt Problem 24 of the 2023 AMC 12A 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 2023 AMC 12A solutions, or check the answer key.

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24.

Let KK be the number of sequences A1,A_1, A2,A_2, ,\ldots, AnA_n such that nn is a positive integer less than or equal to 10,10, each AiA_i is a subset of {1,2,3,,10},\{1,2,3,\ldots,10\}, and Ai1A_{i-1} is a subset of AiA_i for each ii between 22 and n,n, inclusive. For example, {},\{\}, {5,7},\{5,7\}, {2,5,7},\{2,5,7\}, {2,5,7},\{2,5,7\}, {2,5,6,7,9}\{2,5,6,7,9\} is one such sequence, with n=5.n=5. What is the remainder when KK is divided by 10?10?

11

33

55

77

99

Answer: C
Concepts:subsetsmultiplication principlemodular arithmetic
Difficulty rating: 2520
Small Hint:

For a fixed n,n, each of the 1010 elements independently chooses the first set it enters, or never enters

Big Hint:

That gives (n+1)10(n+1)^{10} chains, so K=k=211k10;K=\sum_{k=2}^{11}k^{10}; reduce each term modulo 1010

Solution:

For a fixed length n,n, each element of {1,,10}\{1,\ldots,10\} independently either never appears or first appears in one of A1,,An,A_1,\ldots,A_n, giving n+1n+1 choices. Hence there are (n+1)10(n+1)^{10} chains of length n.n.

Summing, K=n=110(n+1)10=k=211k10. K=\sum_{n=1}^{10}(n+1)^{10}=\sum_{k=2}^{11}k^{10}. Modulo 10,10, the terms k=2,,11k=2,\ldots,11 reduce to 4,9,6,5,6,9,4,1,0,1,4,9,6,5,6,9,4,1,0,1, which sum to 455.45\equiv 5.

Thus, the correct answer is C.

Problem 23#23
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