Consider N=2004^2004.
1) What are the first 3 digits of N?
2) What are the last 3 digits of N?
Another method, requiring a program or spreadsheet, would be working with truncated numbers, or stripping away digits. I would expect this to give the same answer as Charlie's.
Part 1) For the first 3 digits of N,
Start with 2004.
Multiply by 2004 then strip away all but the first 4 digits.
Multiply the resulting 4 digit number by 2004, strip away all but the first 4 digits
we now have a 4 digit number, the first 3 digits of which are the first 3 digits of 2004^3; we've done our procedure twice.
Starting from the beginning and doing the procedure 2003 times will give the first 3 digits of 2004^2004
Part 2) To find the last 3 digits of N, realize that anything in the thousands column or greater would not contribute to the last 3 digits.
Start with 2004.
Multiply by 4. Keep only the last 3 digits.
at this point, we've done our procedure once and we have the last 3 digits of 2004^2
Starting from the beginning and doing the procedure 2003 times will give the last 3 digits of 2004^2004.

Edited to show spreadsheet results:
I got the same last 3 digits as Charlie: 256
But I got different first 3 digits, so at this point I assume I either made a mistake in my formula, or the rounding errors are not as insignificant as I thought: I got 675 for the first 3 digits.
Edited on November 23, 2004, 4:26 pm

It was the rounding. Here is a list of the first several digits based on how many digits I kept during calculations:
4 digits: 6757
5 digits: 95905
6 digits: 100305
7 digits: 1006598
8 digits: 10069507
9 digits: 100698625
10 digits: 1006989853
11 digits: 10069901954 which is the limit of my spreadsheet's precision.
Note Charlie said: "In fact the first several digits of this 6618digit number must be 10069902351."
Edited on November 23, 2004, 4:42 pm

Posted by Larry
on 20041123 15:35:19 