2010/025) For which positive integer values of n occur EXACTLY two 0's at the end of 1^n + 2^n + 3^n + 4^n (base 10)

As it can have maximum 2 zeroes
Let us first find the period at which 1^n+2^n+3^n+ 4^n mod 100 repeats.

The period has to be < 100 as there are maximum number if 100 remainders



1^n mod 100 =1

As 3 is coprime to 100

3^40 = 1 because 40 is for n = 100 as per http://en.wikipedia.org/wiki/Euler%27s_t…

Because 2 and 4 are factors of 100 we need to find period n for which 2^n repeats to base 25. The for 25 is 20

That is 2^20 mod 25 = 1 and 4^20 mod 25 = 1

3 has a period factor of 40( 40 is maximum) and 2 and 4 have a period factor of 20 that is after that \it repeats

So 1^n,2^n, 3^n, 4^n has a period of 40 that is it repeats every 40 times for mod 100

For 1 to 40 we need to find for which the remainder 1^n+2^n+3^n+4^n is 0 mod 1

By a computer program it is found to be 3,5,7,11,15,19,23,25,27,31,35,39.

The values are 40k + 3,40k + 5, 40k + 7, 40k + 11, 40k + 15, 40k + 19, 40k + 23, 40k + 25, 40k + 27, 40k + 31, 40k + 35, 40k + 39 where k is an integer >= 0