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.

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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
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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 6618-digit number must be 10069902351."

Edited on November 23, 2004, 4:42 pm

Posted by Larry on 2004-11-23 15:35:19