What are the last three digits of 49³⁰³ * 3993²⁰² * 39⁶⁰⁶?

The expression can be simplified to: 7^606*11^606*13^606*3^808

Now,

7^606*11^606*13^606*3^808

= (7*11*13)^606*3^808

= (1001)^606*3^808

Now, 1001=1 (mod 1000)
or, 1001^606= 1(mod 1000)

3^800 = 9^400
= (10-1)^400
Expanding this by the binomial theorem, we have:
= M(1000)+1 (where M(p) is a multiple of p)

So, 3^800 = 1(mod 1000)
 
And, 3^8 = 9^4
= (10-1)^4
= (10^4-4*10^3+6*10^2-4*10+1)
=M(1000) + 600-40+1
=M(1000)+561
or, 3^8= 561(mod 1000)

So,
49^303*3993^202*39^606
=7^606*11^606*13^606*3^808
= 1*1*561 (mod 1000) = 561 (mod 1000)

Consequently, the required last three digits are 561. 

Edited on January 16, 2013, 1:46 pm

Edited on January 16, 2013, 2:04 pm

Posted by K Sengupta on 2013-01-16 13:45:49