solve this problem? (number)?

Question

assume A is (1998^1998)^1998
then find out the last 4 digits of A!

Answer 1

8296

Answer 2

It is an Euler's Theorem problem or at least you can work it out like that. And you want the last 4 digits. You first work out 1998^1998 (mod 10,000) the mod 10,000 is because you want the last 4 digits. Then take this answer and do 1998 to the answer you just worked out (mod 10000) and you get your answer. I would work it out but I don't have my calculator on me.

I didn't ever want to see my calculator again after completing a very hard Number Theory exam. :) But I ended up getting 83 % :D

Answer 3

I reckon it is 6016.

After a bit of experimenting with a spreadsheet, I've shown two interesting facts:

1998^4 ends in 6016.
1998^500 ends in 9376.

6016 x 9376 ends in 6016.

So 1998^4, 1998^504, 1998^1004, 1998^1504 and so on all end in 6016.

Now (1998^1998)^1998 is equal to 1998^(1998*1998) = 1998^3992004. Since this exponent ends in 004, the value will end in 6016.

Answer 4

You first!

Answer 5

8296