# Homework Help: 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?