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**