Question

$WhatistheremainderwhenN=(1!+2!+3!+...1000{)}^{40}isdividedby10?$

• A. 1
• B. 3
• C. 7
• D. 8

Answer 1

The correct option is A 1

N = (1! + 2! + 3! + 4! +....1000!)
Now we have to check only
1! + 2! + 3! + 4! as after that will factorial has unit digit as 0

5! = 120
6! = 720 and so on
Thus unit digit of (1! + 2! + 3! + 4! + 0) for all other
$\Rightarrow (1+2+6+4+0{)}^{40}$
Unit digits of all factorial
$\Rightarrow (3{)}^{40}=(3{)}^{4k}$
$\Rightarrow 3\times 3\times 3\times 3\Rightarrow 1$ unit digit. Hence $when(3{)}^{40}\pm 10,Remainderwillbe1.$