Tune my code to solve a puzzle - performance

Question

5 monkey share n peaches, they cannot distribute equally. So the first monkey dump 1 peach, and total number of peaches can be divided by 5, and the first monkey took his part.
Then is the second monkey, -1 peach, can be divided by 5 and took his part.
Until the fifth monkey finished all the steps. There may be some peaches still left.
Give the minimum number of peaches that satisfy this condition.
perl code 1:
#!/usr/bin/perl -w
for $n (0..10000){ #this is basic idea but code is too messy !
if( ($n-1) % 5 == 0 ){
$remain = 4/5 * ($n -1 );
if( ($remain - 1) % 5 == 0){
$remain = 4/5 * ($remain -1 );
if( ($remain - 1) % 5 == 0){
$remain = 4/5 * ($remain -1 );
if( ($remain - 1) % 5 == 0){
$remain = 4/5 * ($remain -1 );
if( ($remain - 1) % 5 == 0){
$remain = 4/5 * ($remain -1 );
print "remain: $remain original: $n\n";
}
}
}
}
}
}
perl code 2:
sub doit($){
($n) = #_;
if( ($n - 1) % 5 ==0 ){ #if can be distributed by 5 monkey
$n = ($n - 1) * 4/5; #commit distribute
return $n;
}else{
return -1; #fail
}
}
for $n (0..10000){ #restriction
$r = $n; #"recursively" find solution
$o = $n; #backup n
$count = 0;
for ($i = 0; $i < 5; $i++){ #assume there is 5 monkey, it can be changed
$r = doit($r);
if($r == -1){ #skip once fail
last;
}
$count++;
}
if($count == 5){ # if pass 5 test, then you found the number !
print "now ".$r."\n";
print "origin ".$o."\n";
}
}
I am thinking to cut some code. But felt hard. Can anyone help ?

Answer 1

First of all, you really should use strict and warnings pragmas at the top of your scripts. Your $n usage is especially worrisome. In the future, if you declare variables with my but use the same name, you convey the fact that they will represent the same quantity, without the fear that they might collide.
Anyway here is a slightly polished, and more importantly strict and warnings safe version:
#!/usr/bin/env perl
use strict;
use warnings;
sub doit {
my ($n) = #_;
if( ($n - 1) % 5 ==0 ){ #if can be distributed by 5 monkey
$n = ($n - 1) * 4/5; #commit distribute
return $n;
} else {
return undef; #fail
}
}
OUTER: for my $n (0..10000){ #restriction
my $r = $n; #"recursively" find solution
for (1..5){ #assume there is 5 monkey, it can be changed
$r = doit($r);
next OUTER unless defined $r;
}
# if code gets here, then it passed 5 test, then you found the number !
print "now: $r\torigin: $n\n";
}
And now, if you really want to be fun with it (don't use this in production, readability first! ):
#!/usr/bin/env perl
use strict;
use warnings;
OUTER: for my $n (0..10000){
my $r = $n;
$r = ($r - 1) % 5 ? next OUTER : 4/5 * ($r - 1) for (1..5);
print "now: $r\torigin: $n\n";
}
or even golfed:
for(0..10000){$r=$n=$_;map$r*=--$r%5?next:4/5,1..5;print"now: $r\torigin: $n\n"}

Answer 2

Consider this solution:
sub share {
($_[0] - 1) % 5 == 0 ? ($_[0]-1)/5*4 : die "unable to share";
}
for my $i (1..10000) {
eval {
share(share(share(share(share($i)))));
};
unless ($#) {
print "solution: $i\n";
last;
}
}
I'm sure there is a monad lurking within.

Answer 3

I'm not 100% sure I understand your question, but instead of searching for the answer, start with the last monkey. The minimum peaches he could take is 1, and even though there could be some left, to get the minimum, assume there are 0 left. Now, calculate how many peaches the second to last monkey saw, and so on.
There is no need to loop, if you start from the last monkey
# set numPeaches to what the last monkey had
$numPeaches = 1;
# then, figure out how many the second to last monkey had, and add to numPeaches
# and, so on ...
# until you get to the first monkey