Hi Diogo. :>) You are so close. A favorite subject of mine. You may
find this of interest.
For your stated problem, your large first number is written in the form of
10^x
When an intermediate number is large (ie 10^32), there is a Number Theory
procedure that is much more efficient by avoiding such a large calculation.
Problems of the form Mod(n^m,x) are usually not calculated like this
directly, but are usually calculated by a function that usually goes by the
name POWERMOD.
If you do an internet search, you should be able to find a number of
different techniques.
To give you some idea, the following calculation would be hard.
81703395 ^ 81703395
It has over 646,000,000 digits in the answer.
Given that, who knows how astronomically large the first number is here:
Mod(123456789012345678901234567 ^ 123456789012345678901234567, 97)
However, Excel can calculate this easily and quickly.
(Mine has been modified to do these types of calculations.)
given that n = 123456789012345678901234567
calculate Mod(n^n,97)
?PowerMod(n, n, 97)
14
So, given that, these are easy.
'Mod(10^29,97)
Debug.Print PowerMod(10, 29, 97)
'Mod(10^33,97)
Debug.Print PowerMod(10, 33, 97)
'Mod(10^999,97)
Debug.Print PowerMod(10, 999, 97)
Returns:
57
28
77
--
Good Luck
HTH
Dana DeLouis