Comments on: FF. Hostile Flowers

By: Philo

Philo — Fri, 13 Mar 2009 01:29:53 +0000

The problem appears to be stated incorrectly (given your solution). You need to start with all the flowers closed to wind up with closed perfect squares, at least, if you’re going to begin by touching them all once.

By: dfollett76

dfollett76 — Sun, 08 Mar 2009 18:08:05 +0000

I’ve spent some time thinking about this before and came up with the wrong answer, this time I tried making an excel spreadsheet that simulated the process and got it right. The key was this function =IF($A3/D$2-INT($A3/D$2)=0,IF(C3=0,1,0),C3)

in this example A3 was the flower number, D2 was the number of the trial (i.e. 3 if every third locker was being opened) ad C3 was the lockers current state.

By: strauss

strauss — Sun, 08 Mar 2009 07:45:16 +0000

The real issue here is:

[spoiler] What characterizes numbers with an odd number of divisors? [/spoiler]

This can be thought of as:

[spoiler] If p is a prime, then how many factors does p^n have?[/spoiler]
[spoiler] If A and B are relatively prime, then the number of factors of A times the number of factors of B = the number of factors of A B (why? Try it out for 2^3 3^3 to see what’s going on) [/spoiler]

By: avgbody

avgbody — Fri, 06 Mar 2009 18:49:53 +0000

[spoiler] With the hint of looking at the first 10, it seems to be that the square numbers are the ones left closed. [/spoiler]

By: 0x616c

0x616c — Thu, 05 Mar 2009 09:01:36 +0000

Nice!
[spoiler]
Any flower whose number is a perfect square is closed.

Not a proof, but to quickly verify:
ruby -e ‘a=100;b=Array.new(a+1,1);1.upto(a){|c|c.step(a,c){|d|b[d]^=1}};1.upto(a){|e|print”flower #{e} is “;puts 1==b[e]?”open”:”closed”}’
[/spoiler]