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Difference between revisions of "2023 AMC 12A Problems/Problem 24"

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Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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==Problem==
  
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-Jonathan Yu
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Let <math>K</math> be the number of sequences <math>A_1</math>, <math>A_2</math>, <math>\dots</math>, <math>A_n</math> such that <math>n</math> is a positive integer less than or equal to <math>10</math>, each <math>A_i</math> is a subset of <math>\{1, 2, 3, \dots, 10\}</math>, and <math>A_{i-1}</math> is a subset of <math>A_i</math> for each <math>i</math> between <math>2</math> and <math>n</math>, inclusive. For example, <math>\{\}</math>, <math>\{5, 7\}</math>, <math>\{2, 5, 7\}</math>, <math>\{2, 5, 7\}</math>, <math>\{2, 5, 6, 7, 9\}</math> is one such sequence, with <math>n = 5</math>.What is the remainder when <math>K</math> is divided by <math>10</math>?
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math>
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==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.
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Thus, the desired value is <math>\sum_{i=1}^{10}(i+1)^10\equiv 5\pmod{10}</math>
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~bluesoul

Revision as of 23:21, 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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