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2024 AMC 10A Problems/Problem 10

Revision as of 01:12, 9 November 2024 by MRENTHUSIASM (talk | contribs)

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 $4 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \cdots$. 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 $100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20 \rightarrow \cdots$. 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 $100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20$. 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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