Problem:
Consider the following operation. Given a positive integer n, if n is a multiple of 3 , then you replace n by 3n​. 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→14→24→8→18→6→2→12→⋯. Suppose you start with n=100. What value results if you perform this operation exactly 100 times?
Answer Choices:
A. 10
B. 20
C. 30
D. 40
E. 50
Solution:
The first several iterations give
100→110→120→40→50→60→20→30→10→20→30→10→⋯
The values will then continue to cycle through 20→30→10. Note that the value 20 occurs after 6,9,12,15,… operations; the value 30 occurs after 7,10,13,16,… operations; and the value 10 occurs after 8,11,14,17,… operations. Because 100 has remainder 1 when divided by 3 , the value after 100 operations is (c)30​.
The problems on this page are the property of the MAA's American Mathematics Competitions