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Difference between revisions of "2023 AMC 12A Problems/Problem 24"
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==Solution 1==
 
==Solution 1==
  
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Consider any sequence with <math>n</math> 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 <math>n</math>th spot, which means every number has <math>(n+1)</math> choices to show up in the sequence. Consequently, for each sequence with length <math>n</math>, there are <math>(n+1)^10</math> possible ways.
+
Consider any sequence with <math>n</math> 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 <math>n</math>th spot, which means every number has <math>(n+1)</math> choices to show up in the sequence. Consequently, for each sequence with length <math>n</math>, there are <math>(n+1)^{10}</math> possible ways.
  
 
Thus, the desired value is <math>\sum_{i=1}^{10}(i+1)^10\equiv 5\pmod{10}</math>
 
Thus, the desired value is <math>\sum_{i=1}^{10}(i+1)^10\equiv 5\pmod{10}</math>
  
 
~bluesoul
 
~bluesoul

Revision as of 22:21, 9 November 2023

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 $\sum_{i=1}^{10}(i+1)^10\equiv 5\pmod{10}$

~bluesoul

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