Finding last two digits for a large exponentiation

1. Nov 2, 2015

1. The problem statement, all variables and given/known data
Find the last two digits in $12345^{6789}$

2. Relevant equations
I reckon solving $12345^{6789} mod(100)$ would give the last two digits.

3. The attempt at a solution
I know that any number that ends with a 5 raised to any positive integer will end with a 5. I also know that the ten digit before the five and the exponent affects the outcome. Both being odd results in 75 as the last two digits and all other odd/even combinations of the two will result in 25 being the last two digits. But I cannot find information or a derivation of why this is the case.

Can someone help me on how to solve this congruence and maybe explain why those rules actually apply? I know the answer but I want to be able to present it with solid mathematics rather than "I read it online".

Thanks

2. Nov 2, 2015

Orodruin

Staff Emeritus
It will be much easier if you start applying some of the basic rules such as a^n = (a-km)^n (mod m), where k is an integer.

3. Nov 3, 2015

Well I know that $12345^2=25 (mod 100)$ but I'm having trouble using this to get on with the calculations. I figure there's some easy trick since the exponent is such a ridiculously large number. How do I proceed (if my findings are meaningful that is)?

4. Nov 3, 2015

Orodruin

Staff Emeritus
So why do you know that? What argumentation led to it?