It's easy to determine how many digits you'll have in the resulting number.

I'm a bit puzzled: sure, logarithms let you replace multiplication with addition -- part of the magic of slide rules -- but isn't taking the base-10 log of something typically as much or more labor intensive then doing the multiplication?

For example: 321 * 311 produces five digits, while 321 * 312 produces six digits.

Sure, you could work out that 10 ** 2.5065 * 10 ** 2.493 = 10 ** 4.9995 and thus five digits, while 10 ** 2.5065 * 10 ** 2.494 = 10 ** 5.0005 and thus six, but is that really easier?

Why not just use something like this:

#!/usr/bin/perl -wl

sub logb10 {
  return log(+shift)/log(10);
}

print int(logb10(100) + logb10(100) + 1);
print int(logb10(321) + logb10(311) + 1);
print int(logb10(321) + logb10(312) + 1);
__DATA__
output:
5
5
6
[download]

Or maybe we could just pull out the magic length method and stop trying to focus on a problem that isn't really a problem?

It probably amounts to much the same thing, but there is a log10() function available as a part of POSIX which is in the standard distribution.

use POSIX qw[log10];

print log10 312;
2.49415459401844
[download]

---

Examine what is said, not who speaks.
"Efficiency is intelligent laziness." -David Dunham
"When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller
If I understand your problem, I can solve it! Of course, the same can be said for you.