Algebra – 28 (CAT-2005)

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)