The remainder obtained when 1! + 2! + 3! + ... + 100! is divided by 12, is

Question

The remainder obtained when 1! + 2! + 3! + ... + 100! is divided by 12, is (A) 7 (B) 6 (C) 8 (D) 9

Tardigrade Admin · Accepted Answer

Let S = 1! + 2! + 3! + 4! + 5! + ... + 100! = 1! + 2! + 3! + (4! + 5! + ... + 100!) = 1 + 2 + 6 + 12λ (∵ H.C.F. of 4! + 5! + ... + 100! is 4!, which is divisible by 12) = 9 + 12 λ ∴ Remainder is 9.

Q. The remainder obtained when $1! + 2! + 3! + ... + 100!$ is divided by $12$ , is

$7$

$6$

$8$

$9$

Let $S = 1! + 2! + 3! + 4! + 5! + ... + 100!$
$= 1! + 2! + 3! + (4! + 5! + ... + 100!)$
$= 1 + 2 + 6 + 12 λ (∵ H . C . F .$ of $4! + 5! + ...$
$+ 100!$ is $4!$ , which is divisible by $12$ )
$= 9 + 12 λ$
$∴$ Remainder is $9$ .

Q. The remainder obtained when 1!+2!+3!+...+100! is divided by 12, is

7

6

8

9

Let S=1!+2!+3!+4!+5!+...+100! =1!+2!+3!+(4!+5!+...+100!) =1+2+6+12λ(∵H.C.F. of 4!+5!+... +100! is 4!, which is divisible by 12) =9+12λ ∴ Remainder is 9.

Q. The remainder obtained when $1! + 2! + 3! + ... + 100!$ is divided by $12$ , is

$7$

$6$

$8$

$9$

Let $S = 1! + 2! + 3! + 4! + 5! + ... + 100!$
$= 1! + 2! + 3! + (4! + 5! + ... + 100!)$
$= 1 + 2 + 6 + 12 λ (∵ H . C . F .$ of $4! + 5! + ...$
$+ 100!$ is $4!$ , which is divisible by $12$ )
$= 9 + 12 λ$
$∴$ Remainder is $9$ .