I'm not sure I understood the last half of your node, but, no, you don't get an "ordinary" increment just because a scalar looks like a number. But it can be hard to distinguish an ordinary (numeric) increment from a magic (stringy) increment in such a case. But here are a couple of examples that demonstrate the difference:

#!/usr/bin/perl -w
use strict;

my $n= '009';
$n++;
print $n, $/;
print 0+$n, $/;
print $n, $/;
$n++;
print $n, $/;
$n= join '', (1..9,0)x7;
$n++;
print $n, $/;
print 0+$n, $/;
print $n, $/;
my $p= $n;
$n++;
print $n, $/;
print $n-$p, $/;
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which produces (minus the comments, of course):

010   # magic increment preserves leading zeros
10    # treating it as a number doesn't give leading zeros
010   # but doesn't strip them from the scalar
11    # but the next increment is numeric not stringy
# stringy increment can deal with /any/ number of digits:
1234567890123456789012345678901234567890123456789012345678901234567891
# numbers only have about 20 significant digits:
1.23456789012346e+069
# treating as a number doesn't destroy the string:
1234567890123456789012345678901234567890123456789012345678901234567891
# but the next increment is just numeric
1.23456789012346e+069
0     # and adding 1 to 1e69 does nothing
[download]

Hope that helps.