Consider N=2004^2004.

1) What are the first 3 digits of N?

2) What are the last 3 digits of N?

N = 2004^2004

Taking log on both sides, we obtain:

log N = 2004*log 2004

= 2004*3.3018977171952

= 6617.003025259 (correct to 9 decimal places)

Accordingly,

N = (10^.003025259)*(10^6617)

= (1.0069902)*(10^6617)

= (100.69902)*(10^6615)

Consequently, the first three digits of N are 100

*Edited on ***October 18, 2007, 9:28 am**