What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 ×

Question

What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + ... + 12 × 12! is divided by 13?
(a) 1 
(b) 12 
(c) 11 
(d) 0

abuc0112 · Answer

Can someone please solve this question

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firas92 · Answer

vsriharsha · Answer

Just start solving in pairs of two, then 3 and so on... 
observe a pattern of, 3!-1, 4!-1....

1! + 2*2! = 5 = 3!-1
1! + 2*2! + 3*3! = 23 = 4!-1

Generalised formula of the series

1*1!+2*2!+3*3!......a*a! is

= (a+1)!-1

So, Total sum = 13!-1/13
Reminder = 12

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