N = 1991*1993*1995*1997*1999*2001*2003.
What is the sum of the hundreds, tens and units digits of N?
+505 Since we need to compute the last 3 digits, we can take the whole expression mod 1000:
\(1991\cdot1993\cdot1995\cdot1997\cdot1999\cdot2001\cdot2003 (\text{mod 1000})\\ =-9\cdot-7\cdot-5\cdot-3\cdot-1\cdot1\cdot3 (\text{mod 1000})\\ =-2835 (\text{mod 1000})\\ =165(\text{mod 1000})\)
The sum of the hundreds, tens, and units digit is \(1+6+5=\boxed{12}\)
