1999 AIME Problems/Problem 7
Problem
There is a set of 1000 switches, each of which has four positions, called , and . When the position of any switch changes, it is only from to , from to , from to , or from to . Initially each switch is in position . The switches are labeled with the 1000 different integers , where , and take on the values . At step i of a 1000-step process, the -th switch is advanced one step, and so are all the other switches whose labels divide the label on the -th switch. After step 1000 has been completed, how many switches will be in position ?
Solution
For each th switch (designated by ), it advances itself only one time at the th step; thereafter, only a switch with larger values will advance the th switch by one step provided divides . Let be the max switch label. To find the divisor multiples in the range of to , we consider the exponents of the number . In general, the divisor-count of must be a multiple of 4 to ensure that a switch is in position A:
We consider the cases where the 3 factors above do not contribute multiples of 4.
- Case of no 2's:
- The switches must be . There are odd integers in to , so we have ways.
- Case of a single 2:
- The switches must be one of or or .
- Since the terms and are three valid choices for the factor above.
- We have ways.
The number of switches in position A is .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.