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Question

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


A

1006

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B

1

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C

0

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D

None of these

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Solution

The correct option is C

0


Given expression is:

n!+(n!+1)+(n!2)+(n!+3).....+(n!2012)
=2013×n!+(12+34.....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.


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