Question

What is the remainder when n! + (n! + 1) + (n! - 2) + (n! + 3) + ..... + (n! - 2012) is divided by 1006 for n = 1006?

• A. 1006
• B. 1
• C. 0
• D. None of these

Answer 1

The correct option is C

0

Given expression is:

$n!+(n!+1)+(n!-2)+(n!+3).....+(n!-2012)$
$=2013\times n!+(1-2+3-4.....-2012)$

$=2013n!+1006\times (-1)$

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.