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 $4 \to 14 \to 24 \to 8 \to 18 \to 6 \to 2 \to 12 \to \cdots$ . 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

Video Solution 1 by Power Solve

https://youtu.be/wamtu7xm0eU

2024 AMC 10A (Problems • Answer Key • Resources)
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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)

Video Solution 1 by Power Solve

See also