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
Solution To Part II of N = 1996^1996
