Let n! = 1*2*3*...n, for integer n ≥ 1. If p = 1!(2*2!) + (3*3!)
+ (4*4!) +.....(10*10!), then p+2 when divided by 11! Leaves a remainder of
_____

(1)10 (2)0 (3)7 (4)1

Solution
follows here:__Solution:__
Apply a
small trick here. 2*2! Can be simplified as (3-1)2!, which can in turn be
simplified as 3!-2!. Now apply this logic to all the terms and then it will get
simplified as follows:

p
= 1!(2*2!) + (3*3!) + (4*4!) +.....(10*10!)

=
(2*2!) + (3*3!) + (4*4!) +.....(10*10!)

=
{(3-1)*2!} + {(4-1)*3!} + {(5-1)*4!} +.....{(11-1)*10!}

=
{(3*2!) – 2!} + {(4*3!) – 3!} + {(5*4!) – 4!} +............{(11*10!) – 10!}

= (3! + 4! + 5!
+....+11!) – (2! + 3! + 4! + .....+10!)

= 11! – 2! =
11! – 2

=> p+2 = 11!
So, if p+2 is
divided with 11!, it leaves a remainder
of zero

**Answer (2)**
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