The remainder obtained when 1! + 2! + ... + 95! is divided by 15 is

Question

The remainder obtained when 1! + 2! + ... + 95! is divided by 15 is (A) 14 (B) 3 (C) 1 (D) 0

Tardigrade Admin · Accepted Answer

Since, 1 !=1,2 !=2,3 !=6,4 !=24,5 !=120 Since, all terms from 5 ! onwards are divisible by 15 and 1 !+2 !+3 !+4 !=33 ∴ The required remainder after dividing by 15 will be 3.

Q. The remainder obtained when $1! + 2! + ... + 95!$ is divided by $15$ is

14

3

1

0

Since, $1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120$
Since, all terms from $5!$ onwards are divisible by 15 and $1! + 2! + 3! + 4! = 33$
$∴$ The required remainder after dividing by 15 will be 3.

Q. The remainder obtained when 1!+2!+...+95! is divided by 15 is

14

3

1

0

Since, 1!=1,2!=2,3!=6,4!=24,5!=120 Since, all terms from 5! onwards are divisible by 15 and 1!+2!+3!+4!=33 ∴ The required remainder after dividing by 15 will be 3.

Q. The remainder obtained when $1! + 2! + ... + 95!$ is divided by $15$ is

Since, $1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120$
Since, all terms from $5!$ onwards are divisible by 15 and $1! + 2! + 3! + 4! = 33$
$∴$ The required remainder after dividing by 15 will be 3.