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 Year by Year (Posted on 2004-11-23)
Consider N=2004^2004.

1) What are the first 3 digits of N?

2) What are the last 3 digits of N?

 See The Solution Submitted by SilverKnight Rating: 2.3333 (6 votes)

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 Solution To Part II | Comment 7 of 8 |
(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
 Posted by K Sengupta on 2007-10-18 06:28:07

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