2024 AMC 10A Problems/Problem 10

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?

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 clarity pls

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

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2024 AMC 10A Problems/Problem 10

Contents

Problem

Solution 1 (fast)

Solution 2 (more explanatory)

See also