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Difference between revisions of "2024 AMC 10A Problems/Problem 10"

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\frac{n}{3}</math>. If <math>n</math> is not a multiple of <math>3</math>, then you replace <math>n</math> by <math>n+10</math>. Then continue this process. For example, beginning with <math>n=4</math>, this procedure gives <math>4 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots</math>. Suppose you start with <math>n=100</math>. What value results if you perform this operation exactly <math>100</math> times?
 
\frac{n}{3}</math>. If <math>n</math> is not a multiple of <math>3</math>, then you replace <math>n</math> by <math>n+10</math>. Then continue this process. For example, beginning with <math>n=4</math>, this procedure gives <math>4 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots</math>. Suppose you start with <math>n=100</math>. What value results if you perform this operation exactly <math>100</math> times?
  
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$\textbf{(A) }10\qquad\
+
$\textbf{(A)}10\qquad\$
  
 
== Solution 1 (Fast Solution) ==
 
== Solution 1 (Fast Solution) ==
 
Let <math>s</math> be the number of times the operation is performed. Notice the sequence goes <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20 \to \cdots</math>. Thus, for <math>s \equiv 1 \pmod{3}</math>, the value is <math>30</math>. Since <math>100 \equiv 1 \pmod{3}</math>, the answer is <math>\boxed{\textbf{(C) }30}</math>.
 
Let <math>s</math> be the number of times the operation is performed. Notice the sequence goes <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20 \to \cdots</math>. Thus, for <math>s \equiv 1 \pmod{3}</math>, the value is <math>30</math>. Since <math>100 \equiv 1 \pmod{3}</math>, the answer is <math>\boxed{\textbf{(C) }30}</math>.
  
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~andliu766
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~andliu766 ~minor fix by MID_HAT
  
 
== Solution 2 (More Explanatory) ==
 
== Solution 2 (More Explanatory) ==

Latest revision as of 14:00, 2 November 2025

Problem

Consider the following operation. Given a positive integer $n$, if $n$ is a multiple of $3$, then you replace $n$ by $\frac{n}{3}$. If $n$ is not a multiple of $3$, then you replace $n$ by $n+10$. Then continue this process. For example, beginning with $n=4$, this procedure gives $4 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots$. Suppose you start with $n=100$. What value results if you perform this operation exactly $100$ times?

$\textbf{(A)}10\qquad$

Solution 1 (Fast Solution)

Let $s$ be the number of times the operation is performed. Notice the sequence goes $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20 \to \cdots$. Thus, for $s \equiv 1 \pmod{3}$, the value is $30$. Since $100 \equiv 1 \pmod{3}$, the answer is $\boxed{\textbf{(C) }30}$.

~andliu766 ~minor fix by MID_HAT

Solution 2 (More Explanatory)

Looking at the first few values of our operation, we get $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20$. We can see that $30$ goes to $10$, then to $20$, then back to $30$, and the loop resets. After 7 operations, we reach $30$. We still have 93 operations left, so because the loop will run exactly $31$ times $(93/3)$, we will reach $30$ again. So, the answer is $\boxed{\textbf{(C) } 30}$.

~Moonwatcher22

Solution 3 (very slightly different than previous)

Calculating the first few values, we get $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20$. We can see that $20$ will go to $30$, then to $10$, then back to $20$, and then the loop resets. After $6$ moves, we reach $20$, the start of the cycle. We still have $100-5$ moves to go, so to find what number we land on after $95$ more steps, we can do $95 \pmod {3} \equiv (9 + 4) \pmod {3} = 1$, meaning we go from $20 \to \boxed{\textbf{(C) } 30}$.

~yuvag

~a lot of credit to Moonwatcher22

Video Solution(Faster!)

https://youtu.be/l3VrUsZkv8I

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/wamtu7xm0eU

Video Solution by Daily Dose of Math

https://youtu.be/17-o9qiprI0

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution by Just Math⚡

https://www.youtube.com/watch?v=lqZUYJPq_Jo

Video Solution by Dr. David

https://youtu.be/Q0LBITGGkGc

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=1547s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
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2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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All AMC 10 Problems and Solutions

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