S = 1 ! + 2 ( 2 ! ) + 3 ( 3 ! ) + ⋯ + 1 0 ( 1 0 ! )
For S as defined above, what is the remainder when S + 2 is divided by 1 1 ! ?
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S = 1 ! + ( 2 − 1 ) 2 ! + ( 3 − 1 ) 3 ! + . . . . . . . . . . . . . . . . . . + ( 1 1 − 1 ) 1 0 !
S = 1 1 ! − 1
S + 2 = 1 1 ! + 1
Therefore remainder * 1
Right approach. Chiranjeev
S can be written as:
S = n = 1 ∑ 1 0 n ( n ! ) = n = 1 ∑ 1 0 ( ( n + 1 ) n ! − n ! ) = n = 1 ∑ 1 0 ( n + 1 ) ! − n = 1 ∑ 1 0 n ! = 1 1 ! − 1 !
Therefore, S + 2 ≡ 1 1 ! − 1 ! + 2 ≡ 1 (mod 11) .
We know that, n × n ! = ( 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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Note that, n × n ! = ( n + 1 − 1 ) n ! = ( n + 1 ) ! − n ! S = i = 1 ∑ 1 0 n × n !
Telescoping S with the help of above result,
we get S = 1 1 ! − 1 (How)
⇒ S + 2 = 1 1 ! + 1 Clearly remainder is 1