2023 AMC 12A Problems/Problem 24

Problem

Let $K$ be the number of sequences $A_1$ , $A_2$ , $\dots$ , $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ , $\{5, 7\}$ , $\{2, 5, 7\}$ , $\{2, 5, 7\}$ , $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$ ?

$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution 1

Consider any sequence with $n$ terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the $n$ th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence with length $n$ , there are $(n+1)^{10}$ possible ways.

Thus, the desired value is $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{\textbf{(C) } 5}\pmod{10}$

~bluesoul

Solution 2

Let $f(x,\ell)$ be the number of sequences $A_1$ , $A_2$ , $\dots$ , $A_\ell$ such that each $A_i$ is a subset of $\{1, 2,\dots,x\}$ , and $A_i$ is a subset of $A_{i+1}$ for $i=1$ , $2\dots$ , $\ell-1$ . Then $f(x,1)=2^x$ and $f(0,\ell)=1$ .

If $\ell\ge2$ and $x\ge1$ , we need to get a recursive formula for $f(x,\ell)$ : If $|A_1|=i$ , then $A_1$ has $\text C_x^i$ possibilities, and the subsequence $\{A_i\}_{2\le i\le\ell}$ has $f(x-i,\ell-1)$ possibilities. Hence $f(x,\ell)=$ $f(x,\ell)=\sum_{i=0}^x\text C_x^if(x-i,\ell-1).$ By applying this formula and only considering modulo $10$ , we get $f(1,2)=3$ , $f(1,3)=4$ , $f(1,4)=5$ , $f(1,5)=6$ , $f(1,6)=7$ , $f(1,7)=8$ , $f(1,8)=9$ , $f(1,9)=0$ , $f(1,10)=1$ , $f(2,2)=9$ , $f(2,3)=6$ , $f(2,4)=5$ , $f(2,5)=6$ , $f(2,6)=9$ , $f(2,7)=4$ , $f(2,8)=1$ , $f(2,9)=0$ , $f(2,10)=1$ , $f(3,2)=7$ , $f(3,3)=4$ , $f(3,4)=5$ , $f(3,5)=6$ , $f(3,6)=3$ , $f(3,7)=2$ , $f(3,8)=9$ , $f(3,9)=0$ , $f(3,10)=1$ , $f(4,2)=1$ , $f(4,3)=6$ , $f(4,4)=5$ , $f(4,5)=6$ , $f(4,6)=1$ , $f(4,7)=6$ , $f(4,8)=1$ , $f(4,9)=0$ , $f(4,10)=1$ , $f(5,2)=3$ , $f(5,3)=4$ , $f(5,4)=5$ , $f(5,5)=6$ , $f(5,6)=7$ , $f(5,7)=8$ , $f(5,8)=9$ , $f(5,9)=0$ , $f(5,10)=1$ , $f(6,2)=9$ , $f(6,3)=6$ , $f(6,4)=5$ , $f(6,5)=6$ , $f(6,6)=9$ , $f(6,7)=4$ , $f(6,8)=1$ , $f(6,9)=0$ , $f(6,10)=1$ , $f(7,2)=7$ , $f(7,3)=4$ , $f(7,4)=5$ , $f(7,5)=6$ , $f(7,6)=3$ , $f(7,7)=2$ , $f(7,8)=9$ , $f(7,9)=0$ , $f(7,10)=1$ , $f(8,2)=1$ , $f(8,3)=6$ , $f(8,4)=5$ , $f(8,5)=6$ , $f(8,6)=1$ , $f(8,7)=6$ , $f(8,8)=1$ , $f(8,9)=0$ , $f(8,10)=1$ , $f(9,2)=3$ , $f(9,3)=4$ , $f(9,4)=5$ , $f(9,5)=6$ , $f(9,6)=7$ , $f(9,7)=8$ , $f(9,8)=9$ , $f(9,9)=0$ , $f(9,10)=1$ , $f(10,2)=9$ , $f(10,3)=6$ , $f(10,4)=5$ , $f(10,5)=6$ , $f(10,6)=9$ , $f(10,7)=4$ , $f(10,8)=1$ , $f(10,9)=0$ , $f(10,10)=1$ .

Lastly, we get $K\equiv\textstyle\sum\limits_{i=1}^{10}f(10,i)\equiv\boxed{\textbf{(C) } 5}\pmod{10}$ . ~Quantum-Phantom

Solution 3 (Cheese)

Since the question only wants mod 10 of the answer, we can cheese this problem. Let $b_n$ be the number of elements of the set $a_n$ . Assume that $b_k\neq0$ and $b_k\neq10$ for any $k$ . However, that means by symmetry, there will be $\binom{10}{k}$ different sequences of $a$ with the same sequence of $b$ . Since $\binom{10}{k}$ is 0 mod 10 for all $k$ except for 0 and 10, we only consider sequences where each term is either the empty set or the entire set ${1,2,\cdots,10}$ . If $n=1$ , then there are $2$ sets. If $n=2$ , then there are $3$ sets, and so on. So the answer is $2+3+\cdots{}+11=65==5$ (mod 10).

Video Solution 1 by OmegaLearn

https://youtu.be/0LLQW0XCKsQ

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 12 Problems and Solutions

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