What is the remainder when n! + (n! + 1) + (n! - 2) + (n! + 3) + ..... + (n! - 2012) is divided by 1006 for n = 1006?
0
Given expression is:
n!+(n!+1)+(n!−2)+(n!+3).....+(n!−2012)
=2013×n!+(1−2+3−4.....−2012)
=2013 n!+1006×(−1)
Now for n = 1006, the expression 2013 n! - 1006 is clearly divisible by 1006.
Hence the remainder is zero.
Shortcut Method:
The answer will be same for n = 1, n = 2 and so on.. We can just substitute one of these values and find the answer. For instance, when n = 2:
The series is: 2 + (2+1) + (2-2) + (2+3) + (2-4) = 8 which when divided by 2 gives a remainder 0.