Do you love factorial?

S = 1 ! + 2 ( 2 ! ) + 3 ( 3 ! ) + + 10 ( 10 ! ) \large S=1!+2(2!)+3(3!)+\cdots+10(10!)

For S S as defined above, what is the remainder when S + 2 S+2 is divided by 11 ! 11! ?


You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.


The answer is 1.

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4 solutions

Sanjeet Raria
Oct 1, 2014

Note that, n × n ! = ( n + 1 1 ) n ! = ( n + 1 ) ! n ! n×n!=(n+1-1)n!=(n+1)!-n! S = i = 1 10 n × n ! S=\sum_{i=1}^{10}n×n!

Telescoping S S with the help of above result,

we get S = 11 ! 1 S=11!-1 (How)

S + 2 = 11 ! + 1 \Rightarrow S+2=11!+1 Clearly remainder is 1 \textbf {Clearly remainder is 1}

overrated ._.

math man - 6 years, 8 months ago
U Z
Oct 1, 2014

S = 1 ! + ( 2 1 ) 2 ! + ( 3 1 ) 3 ! + . . . . . . . . . . . . . . . . . . + ( 11 1 ) 10 ! S = 1! + (2-1)2! + (3 - 1)3! + .................. + (11 -1)10!

S = 11 ! 1 S = 11! -1

S + 2 = 11 ! + 1 S +2 = 11! +1

Therefore remainder * 1

Right approach. Chiranjeev

Chiranjeev Mishra - 6 years, 6 months ago
Chew-Seong Cheong
Sep 13, 2018

S S can be written as:

S = n = 1 10 n ( n ! ) = n = 1 10 ( ( n + 1 ) n ! n ! ) = n = 1 10 ( n + 1 ) ! n = 1 10 n ! = 11 ! 1 ! \begin{aligned} S & = \sum_{n=1}^{10} n(n!) \\ & = \sum_{n=1}^{10}\big((n+1)n! - n!\big) \\ & = \sum_{n=1}^{10} (n+1)! - \sum_{n=1}^{10} n! \\ & = 11! - 1! \end{aligned}

Therefore, S + 2 11 ! 1 ! + 2 1 (mod 11) S+2 \equiv 11!-1!+2 \equiv \boxed 1 \text{ (mod 11)} .

Somoy Subandhu
Nov 2, 2014

We know that, n × n ! = ( n + 1 1 ) n ! = ( n + 1 ) ! n ! n\quad\times\quad n!=\quad(n+1-1)n!=(n+1)!-n! now let's see the given series

S=1!+2.2!+3.3!+........+10.10!

seeing the formula above we can say that,

S= 1! + (2+1)!-2!+(3+1)!-3!+.......+(10+1)!-10!

or, S=1!+3!-2!+4!-3!+5!-4!+6!-5!+7!-6!+8!-7!+9!-8!+10!-9!+11!-10!

S=1!-2!+3!-3!+4!-4!+........+10!-10!+11!

so, S=11!-2!+1! = 11!-2+1 = 11!-1

now adding 2 on both sides we get , S+2 = 11!+1

As 11!=1.2.3........11 , which is divisible by 11 , hence the remainder is 1

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