Even though is a huge number, it still ends in 001.
Therefore, if I am not mistaken, will end in 321.
I am trying to find a way to obtain the last 3 digits of k,
where k = 4321^(10^4321 + 1).
I can see that 10^4321 + 1 = log(k) / log(4321),
or log(k) / log(4321) - 10^4321 = 1
However, numbers are so huge here that I think the use of
some trick or theory is required.
Can anyone help. Thank you.
Hello, Wilmer!
I used a very primtive approach . . .Find the last 3 digits of , where: .
Since we are concerned with the last three digits only,
. . we can raise to various powers.
I found that: . ends in
That is, consecutive powers of 321 end in: .
. . It goes through a 25-step cycle and starts over.
The power
So we have: .
What we want is
We need to use Euler's totient theorem mod n
is the number of numbers smaller than 1000 and coprime with 1000.
1000 =
so we just need to count the number of multiples of 2 and 5 less than 1000. There are 1000/2 = 500 even numbers, 1000/5 = 200 multiples of 5 and 1000/10=100 multiples of 2 and 5.
This gives us = 1000-500-200+100 = 400
so we have
but 400|
so =321