Question

Find the last three digits of the sum A=11!+12!+13!+...+2006!​

Answer 1

Let’s see what happens to the last three digits
11! = ...800
12! = ...600
13! = ...800
14! = ...200
15! = ...000
Every remaining factorial will end in 000
800 + 600 + 800 + 200 = 2400
So the last three digits of 11! + 12! + 13! + ... + 2006! are 400

Answer 2

996

Explanation:

n = 2006-11+1 = 1996

the sum A = 1996/2 × (11+2006)

= 2,012,996

so, the last 3 digits : 996