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

(→‎Solution 3 (very slightly different than previous))
(→‎Video Solution by jmath⚡)
 
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<math>\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50</math>
 
<math>\textbf{(A) }10\qquad\textbf{(B) }20\qquad\textbf{(C) }30\qquad\textbf{(D) }40\qquad\textbf{(E) }50</math>
  
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== Solution 1 (fast ⚡️⚡️⚡️) ==
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== 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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== Solution 2 (More Explanatory) ==
 
== Solution 2 (More Explanatory) ==
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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>.
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edit for grammar pls
 
  
 
~Moonwatcher22
 
~Moonwatcher22
  
 
== Solution 3 (very slightly different than previous) ==
 
== Solution 3 (very slightly different than previous) ==
−
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>20</math> will go to <math>30</math>, then to <math>10</math>, then back to <math>20</math>, and the loop resets. After <math>6</math> operations, we reach <math>20</math>, the start of the cycle. We still have <math>100-6</math> operations left, so to find what 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 </math>\boxed{\textbf{(C) } 30}<math></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
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~alot of credit to Moonwatcher22
 
~alot of credit to Moonwatcher22
  
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== Video Solution 1 by Power Solve ==
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== Video Solution by Pi Academy ==
 +
 
 +
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
 +
 
 +
==Video Solution 1 by Power Solve ==
  
 
https://youtu.be/wamtu7xm0eU
 
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
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==Video Solution by Just Math⚡==
 +
https://www.youtube.com/watch?v=lqZUYJPq_Jo
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:38, 17 November 2024

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

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

~alot of credit to Moonwatcher22

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

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 10 Problems and Solutions

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