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

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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 <math>n</math>, if <math>n</math> is a multiple of <math>3</math>, then you replace <math>n</math> by <math>
−
\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 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \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?
  
 
<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>
  
 
== Solution 1 (fast ⚡️⚡️⚡️) ==
 
== Solution 1 (fast ⚡️⚡️⚡️) ==
−
Let <math>s</math> be the number of times the operation is performed. Notice the sequence goes <math>100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20 \rightarrow \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
  
 
== Solution 2 (More Explanatory) ==
 
== Solution 2 (More Explanatory) ==
−
Looking at the first few values of our operation, we get <math>100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 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> 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>.
  
 
edit for grammar pls
 
edit for grammar pls

Revision as of 02:28, 9 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 ⚡️⚡️⚡️)

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 $30$ will go 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 at $30$ again. So, the answer is $\boxed{\textbf{(C) } 30}$.

edit for grammar pls

~Moonwatcher22

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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