2024 AMC 10A Problem 10

Attempt Problem 10 of the 2024 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

Consider the following operation. Given a positive integer n,n, if nn is a multiple of 3,3, then you replace nn by n3.\tfrac{n}{3}. If nn is not a multiple of 3,3, then you replace nn by n+10.n + 10. Then continue this process. For example, beginning with n=4,n = 4, this procedure gives 4→14→24→84 \to 14 \to 24 \to 8 →18→6→2→12→⋯ .\to 18 \to 6 \to 2 \to 12 \to \cdots.
Suppose you start with n=100.n = 100. What value results if you perform this operation exactly 100100 times?
1010
2020
3030
4040
5050
Answer: C
Concepts:recursionmodular arithmeticprocess simulationpattern recognition
Small hint:
Compute the first several terms from 100;100; the values soon start repeating
Big hint:
Once the sequence reaches 1010 it cycles 10→20→3010 \to 20 \to 30 with period 3;3; locate step 100100 in the cycle
Solution:
Just run it from 100:100: 100→110→120→40100 \to 110 \to 120 \to 40 →50→60→20\to 50 \to 60 \to 20 →30→10→20\to 30 \to 10 \to 20 →30→10→⋯ .\to 30 \to 10 \to \cdots. After the 88th step we’re at 10,10, and from there it cycles 10,20,3010, 20, 30 with period 3.3. So step 8+k8 + k is the kkth entry of the cycle. For step 100,100, k=92,k = 92, and 92≡2(mod3),92 \equiv 2 \pmod 3, which lands on 30.30. Therefore, the answer is C.
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