Frid....err...Monday math solution

The answer to last week's "Lockers" problem: only perfect squares stay open. This can be seen with experimentation. The reason why this is true is due to parity - the number of divisors of n is odd iff n is a perfect square.

Daniel, of course, saw this as an exercise in programming, and wrote a Perl script.

#!/usr/bin/perl
use strict;
use warnings;
my $CLOSED = q{.};
my $OPENED = q{O};
my $NUM_LOCKERS = 80;
my %lockers;
# Set all the lockers to be cloesd
map { $lockers{$_} = $CLOSED } 1 .. $NUM_LOCKERS;
for my $n ( 2 .. $NUM_LOCKERS ){
for my $i ( $n .. $NUM_LOCKERS ) {
# Every nth locked => divisible by n
if ( $i % $n == 0 ) {
# Toggle the state of the locker
$lockers{$i} = $lockers{$i} eq $OPENED ? $CLOSED : $OPENED;
}
}
print "$n:\t", join ( q{}, @lockers{1..$NUM_LOCKERS} ), "\n";
}

I didn't get a chance to blog this on Friday, on account of my having a life. Sorry. The weekly math problem is turning out to be too much of a commitment, it seems (regardless of how helpful these may prove during the occasional job interview). I'll post problems as they come up, but I don't think that I can keep to a schedule.