Question

Consider the following operation. Given a positive integer $n,$ if $n$ is a multiple of $3,$ then you replace $n$ by $\tfrac{n}{3}.$ 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?

$10$

$20$

$30$

$40$

$50$

Solution:

Just run it from $100:$ $100 \to 110 \to 120 \to 40 \to 50 \to 60 \to 20 \to 30 \to 10 \to 20 \to 30 \to 10 \to \cdots.$ After the $8$ th step we're at $10,$ and from there it cycles $10, 20, 30$ with period $3.$ So step $8 + k$ is the $k$ th entry of the cycle. For step $100,$ $k = 92,$ and $92 \equiv 2 \pmod 3,$ which lands on $30.$ Therefore, the answer is C.

2024 AMC 10A Problem 10

Solution:

Problem 10 in Other Years