Question

The tens digit of $1!+2!+3!+\dots +49!$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Given sum is, $1!+2!+3!+\dots +49!$

On evaluation

of the first nine terms we get,

$$\begin{gathered} 1!=1 \\ 2!=2\times 1!=2 \\ 3!=3\times 2!=6 \\ 4!=4\times 3!=24 \\ 5!=5\times 4!=120 \\ 6!=6\times 5!=720 \\ 7!=7\times 6!=5040 \\ 8!=8\times 7!=40320 \\ 9!=9\times 8!=362880 \end{gathered}$$

The tens and units place of $10!$ is $0$ because in the factors of $10!$, we observe that $2$, $5$ and $10$ are present which when multiplied gives $100$. Thus, $100$ is a factor of $10!$.

We know that all factors of $100$ end as $00$ i.e., $0$ at the tens and units places.

We also know that $n!=n\times \left(n-1\right)!$

Thus, all factorials after $10!$ contain a factor of $100$. Therefore, they all have $0$ in their tens and units places.

Thus, to calculate the value of the tens place of the given sum, we only need to take the sum of the last two places of the first nine terms i.e., of $1!$ to $9!$.

Observing the calculations above,

The sum we need to calculate as reasoned above is,

$1+2+6+24+20+20+40+20+80=213$

The tens digit of this sum is $1$.

Thus, the tens digit of the given sum is also $1$

Hence, option A is correct.