Author |
Topic: Circular Jail Cell (Read 3359 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
Circular Jail Cell
« on: Jul 28th, 2002, 6:50am » |
 Quote  Modify
|
Ten prisoners found their door open after 100 rounds of the drunken jailor. They were the ones whose cell number is a perfect square.
Each door is opened or closed once for each factor of its cell number, including 1 and the cell number itself. For example, for cell 12: 1, 2, 3, 4, 6, 12.
Most whole numbers have an even number of factors. This is because each factor f can be paired with n/f. The exceptions are the perfect squares, where f = n/f for precisely one factor.
|
|
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
horseisahorse
Guest
|
Actually, I think the real-world answer would be zero, since each prisoner would have booked out of the prison as soon as the jailor opened his door during the first round.
|
|
IP Logged |
|
|
|
Ivan
Guest
|
I got the same answer as NickH. Since I like doing things the hard way, I wrote a python program (the formatting here ain't great, look at it at http://linuxhelp.hn.org/jail-cell.py if you're interested. (with proper block spacing)):
Code:
#!/usr/bin/python
doors = {}
# create doors 1 - 100
for x in xrange(101):
doors[x] = 0
del doors[0]
# first, the jailer goes to each cell
# this number will be increased in each round
sep = 1
# repeat until jailer falls down
while 1:
doorcount = 1
for x in doors:
# check if the door by the current door skipping factor
if doorcount % sep:
if doors[x] == 0:
# if it's closed, open it
doors[x] = 1
else:
# if it's open, close it
doors[x] = 0
doorcount += 1
sep += 1
if sep > 100:
# jailer falls down
break
# count how many doors are open in the end
doors_open = 0
for x in doors:
if doors[x] == 1:
doors_open += 1
print doors_open
|
|
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print
Remove