1995 AMC 12 Problem 15

Question

Five points on a circle are numbered

1,

2,

3,

4,

and

5

in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point

5,

after

1995

jumps it will be on point

1

2

3

4

5

Answer: D

Concepts:cycles modular arithmetic

Difficulty rating: 1470

Small hint:

Write down the landing points for the first few jumps

Big hint:

Once a point repeats, reduce the remaining number of jumps modulo the cycle length

Solution:

Starting at

5,

the landing points are

1,2,4,1,2,4,\ldots

Thus the cycle

1,2,4

has length

3.

Since

1995

is divisible by

3,

the

1995

th landing point is

4.

Thus the correct answer is D.