The tensdigit of 1 !+2 !+3 !+ ldots+49 ! is

Question

The tensdigit of 1 !+2 !+3 !+ ldots+49 ! is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

We know that 1 !+2 !+3 !+4 !=33. Also, 5 !=120,6 !=720,7 !=5040,8 !=40320 and 9 !=362880. Thus, tens digit of 1 !+2 !+ ldots+9 ! is 1 . Also, note that n ! is divisible by 100 for all n ≥ 10. Therefore, the tens digit of 1 !+2 !+ ldots+49 ! is 1

Q. The tensdigit of $1! + 2! + 3! + \dots + 49!$ is

1

2

3

4

We know that $1! + 2! + 3! + 4! = 33$ .
Also, $5! = 120, 6! = 720, 7! = 5040, 8! = 40320$ and $9! = 362880$ .
Thus, tens digit of $1! + 2! + \dots + 9!$ is $1$ .
Also, note that $n!$ is divisible by $100$ for all $n \geq 10$ .
Therefore, the tens digit of $1! + 2! + \dots + 49!$ is $1$

Q. The tensdigit of 1!+2!+3!+…+49! is

1

2

3

4

We know that 1!+2!+3!+4!=33. Also, 5!=120,6!=720,7!=5040,8!=40320 and 9!=362880. Thus, tens digit of 1!+2!+…+9! is 1 . Also, note that n! is divisible by 100 for all n≥10. Therefore, the tens digit of 1!+2!+…+49! is 1

Q. The tensdigit of $1! + 2! + 3! + \dots + 49!$ is

We know that $1! + 2! + 3! + 4! = 33$ .
Also, $5! = 120, 6! = 720, 7! = 5040, 8! = 40320$ and $9! = 362880$ .
Thus, tens digit of $1! + 2! + \dots + 9!$ is $1$ .
Also, note that $n!$ is divisible by $100$ for all $n \geq 10$ .
Therefore, the tens digit of $1! + 2! + \dots + 49!$ is $1$