Consider N=2004^2004.
1) What are the first 3 digits of N?
2) What are the last 3 digits of N?
(In reply to
Solution To Part I by K Sengupta)
It can easily be shown in terms of analytical methods that:
376^m (mod 1000)
= 376, for any positive integer m............(i)
Now, we know that:
4^10 (Mod 1000) = 576, while:
576^5 (Mod 1000) = 376
Or, 4^50 (Mod 1000) = 376, for any positive integer m (from (i))
........... (ii)
Now, 2004(Mod 1000) = 4
or, 2004^p (Mod 1000) = 4^p (mod 1000), for any non negative integer p.
Thus, substituting m = 40 in (ii), we obtain:
2004^2000 (Mod 1000)
= 4^2000 (Mod 1000)
= 376
But,
4^2002(Mod 1000) = 376*16 (Mod 1000)
= 16
Accordingly,
2004^2004 (Mod 1000)
= 16*16 (Mod 1000)
= 256
Consequently, the last three digits of N are 256
Edited on October 18, 2007, 6:29 am
Solution To Part II
