hmm, I'll start it up a bit, since it is kinda hard to see sometimes how to approach a problem like that...
let's take things mod 100. that is
509485034958 = 58 (mod 100) , so any number mod 100 is its last 2 digits
For the record, a is congruent to b modulo n if and only if n divides a-b, or to put it another way, a-b=n*k, where k is some natural number.
So, for example, 124 is equal to 24 modulo 100
124=24 (mod 100)
Why ? Because 100 divides (124-24)
So,
all we need now is
3^1999 = ?? (mod 100)
Now, how do we do that ?
hmm, well, that's a bit more complicated, but still if you know what to do, it's not that bad.
let me think about it now ... |
