 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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) Comments: ( Back to comment list | You must be logged in to post comments.) Solution To Part II of N = 1996^1996 Comment 8 of 8 | (In reply to Solution To Part II by K Sengupta)

The location of the official solution as referred to by the proposer
has since shifted to this location:

http://www.stetson.edu/mathcs/events/mathcontest/1996/index.shtml#top

The said location envisages obtaining the first three digits and the last three digits of N = 1996^1996.

While the methodology for Part I as posited in this location is quite staightforward, the solution to Part II in terms of our methodology may not be as simple as in Part-I.

For Part-II of this problem, we also need to find the residue of
576^4 when reduced Mod 1000.

Now, 576^2(Mod 1000) = 776
Or, 576^4(Mod 1000) = 776^2(Mod 100) = 176.

Substituting m=39 in the relationship: 376^m (mod 1000) = 376, for any positive integer m, we obtain:

4^1950(Mod 1000) = 376
Now, 4^40(Mod 1000) = 576^4(Mod 1000) = 176.

Thus, 4^1990(Mod 1000) = 376*176 (Mod 1000) = 176
Also, 4^6(Mod 1000) = 4096(Mod 1000) = 96

Since 1996 = -4(Mod 1000), we must have:
1996^1996(Mod 1000) = 4^1996(Mod 1000)
= 176*96(Mod 1000)
= 896

Consequently, the required last three digits of 1996^1996 are 896

Edited on October 18, 2007, 9:21 am
 Posted by K Sengupta on 2007-10-18 06:31:33 Please log in:
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