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