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2024 AMC 10A Problems/Problem 10: Difference between revisions

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== Problem ==
== Problem ==


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


<math>\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50</math>
<imath>\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50</imath>


== Solution 1 (fast ⚡️⚡️⚡️) ==
== 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>.


~andliu766
~andliu766 ~minor fix by MID_HAT


== Solution 2 (More Explanatory) ==
== Solution 2 (More Explanatory) ==
Looking at the first few values of our operation, we get <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20</math>. We can see that <math>30</math> will go to <math>10</math>, then to <math>20</math>, then back to <math>30</math>, and the loop resets. After 7 operations, we reach <math>30</math>. We still have 93 operations left, so because the loop will run exactly <math>31</math> times <math>(93/3)</math>, we will reach at <math>30</math> again. So, the answer is <math>\boxed{\textbf{(C) } 30}</math>.
Looking at the first few values of our operation, we get <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20</math>. We can see that <math>30</math> goes to <math>10</math>, then to <math>20</math>, then back to <math>30</math>, and the loop resets. After 7 operations, we reach <math>30</math>. We still have 93 operations left, so because the loop will run exactly <math>31</math> times <math>(93/3)</math>, we will reach <math>30</math> again. So, the answer is <math>\boxed{\textbf{(C) } 30}</math>.
 
edit for grammar pls


~Moonwatcher22
~Moonwatcher22


== Solution 3 (very slightly different than previous) ==
== Solution 3 (very slightly different than previous) ==
Calculating the first few values, we get <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20</math>. We can see that <math>20</math> will go to <math>30</math>, then to <math>10</math>, then back to <math>20</math>, and then the loop resets. After <math>6</math> moves, we reach <math>20</math>, the start of the cycle. We still have <math>100-6</math> moves to go, so to find what number we land on after <math>94</math> more steps, we can do  <math>94 \pmod {3} \equiv (9 + 4) \pmod {3} = 1</math>, meaning we go from <math>20 \to \boxed{\textbf{(C) } 30}</math>.
Calculating the first few values, we get <math>100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20</math>. We can see that <math>20</math> will go to <math>30</math>, then to <math>10</math>, then back to <math>20</math>, and then the loop resets. After <math>6</math> moves, we reach <math>20</math>, the start of the cycle. We still have <math>100-5</math> moves to go, so to find what number we land on after <math>95</math> more steps, we can do  <math>95 \pmod {3} \equiv (9 + 4) \pmod {3} = 1</math>, meaning we go from <math>20 \to \boxed{\textbf{(C) } 30}</math>.


~yuvag
~yuvag


~alot of credit to Moonwatcher22
~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 ==
==Video Solution 1 by Power Solve ==


https://youtu.be/wamtu7xm0eU
https://youtu.be/wamtu7xm0eU


==See also==
== Video Solution by Daily Dose of Math ==
{{AMC10 box|year=2024|ab=A|num-b=9|num-a=11}}
 
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==
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
* [[AMC 10]]
* [[AMC 10 Problems and Solutions]]
* [[Mathematics competitions]]
* [[Mathematics competition resources]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 20:31, 11 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?

$\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50$

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)
Preceded by
2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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All AMC 10 Problems and Solutions

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