Dillesh4096 wrote:
If x = 1x1! + 2x2! + 3x3! + . . . . . . + 20x20!. What is the remainder when x+2 is divided by 21!
A. 0
B. 1
C. 2
D. 19
E. 20
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I am sure the question is meant to be ..
\(x = 1*1! + 2*2! + 3*3! + . . . . . . + 20*20!\)
\(x = 1*1! + 2*2! + 3*3! + . . . . . . + 20*20!\)
= \((2-1)*1! + (3-1)*2! + (4-1)*3! + . . . . . . + (21-1)*20!\)
=\(2*1!-1!+3*2!-2!+4*3!-3!+.....21*20!-20!\)
=\(2!-1!+3!-2!+4!-3!+......20!-19!+21!-20!\)
=\(21!-1!\)
So \(x+2=21!-1+2=21!+1..\)
when you divide 21!+1 by 21!, the remainder is 1..
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