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 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**