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

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~Quantum-Phantom
 
~Quantum-Phantom
  
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==Solution 3 (Cheese) ==
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==Solution 3 (Cheese, answer by coincidence) ==
 
Since the question only wants mod 10 of the answer, we can cheese this problem.
 
Since the question only wants mod 10 of the answer, we can cheese this problem.
 
Let <math>b_n</math> be the number of elements of the set <math>a_n</math>. Assume that <math>b_k\neq0</math> and <math>b_k\neq10</math> for any <math>k</math>. However, that means by symmetry, there will be <math>\binom{10}{k}</math> different sequences of <math>a</math> with the same sequence of <math>b</math>. Since <math>\binom{10}{k}</math> is 0 mod 10 for all <math>k</math> except for 0 and 10, we only consider sequences where each term is either the empty set or the entire set <math>{1,2,\cdots,10}</math>. If <math>n=1</math>, then there are <math>2</math> sets. If <math>n=2</math>, then there are <math>3</math> sets, and so on. So the answer is <math>2+3+\cdots{}+11=65==5</math> (mod 10).
 
Let <math>b_n</math> be the number of elements of the set <math>a_n</math>. Assume that <math>b_k\neq0</math> and <math>b_k\neq10</math> for any <math>k</math>. However, that means by symmetry, there will be <math>\binom{10}{k}</math> different sequences of <math>a</math> with the same sequence of <math>b</math>. Since <math>\binom{10}{k}</math> is 0 mod 10 for all <math>k</math> except for 0 and 10, we only consider sequences where each term is either the empty set or the entire set <math>{1,2,\cdots,10}</math>. If <math>n=1</math>, then there are <math>2</math> sets. If <math>n=2</math>, then there are <math>3</math> sets, and so on. So the answer is <math>2+3+\cdots{}+11=65==5</math> (mod 10).
 +
 +
*Note: Unfortunately, <math>\binom{10}{5}</math> is not congruent to 5 mod 10, so this solution has the correct answer by coincidence.
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==

Revision as of 16:30, 10 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 $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{\textbf{(C) } 5}\pmod{10}$

~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, answer by coincidence)

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).

  • Note: Unfortunately, $\binom{10}{5}$ is not congruent to 5 mod 10, so this solution has the correct answer by coincidence.

Video Solution 1 by OmegaLearn

https://youtu.be/0LLQW0XCKsQ


See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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