Question

The unit digit of 1!+2!+3!+4!....................+49! is:

(a)1
(b)2
(c)3
(d)4

Answer 1

Hint: Focus on the point that the numbers with 10 as one of its factors have 0 as its unit digit. Also, use the definition of factorial to get the result.

Complete step-by-step answer:

The given expression is: 1!+2!+3!+4!....................+49!, and we are asked to find out the unit digit of the given expression.

We know that $n!=1\times 2\times 3\times .....\times n=\prod\limits_{r=1}^{n}{r}$ . Using this to show that 5! has 10 as one of its factors.

$5!=5\times 4\times 3\times 2\times 1=10\times 12$

So, we can also say that the factorials of the natural numbers greater than or equal to 5 have 10 as one of their factors, and we know that the numbers having 10 as one of their factors is having 0 as its unit digit.

Therefore, we can say that the unit digit of the expression 1!+2!+3!+4!, is the same as the unit digit of the expression 1!+2!+3!+4!....................+49!.

Now we will find out the unit digit of 1!+2!+3!+4!. As the expression has only four terms, we will directly calculate each firm's value and sum it to get the unit digit.

$1!+2!+3!+4!=1+2+6+24=33$

Therefore, the unit digit of 1!+2!+3!+4! Is 3. Hence, the unit digit of 1!+2!+3!+4!....................+49! Is 3 and the answer to the above question is option (c).


Note: The unit digit for $\sum\limits_{r=1}^{n}{r!}$ is always 3, provided n is greater than or equal to 5. It is prescribed that whenever you are asked about the unit digit of expression, always try to avoid solving for a complete answer and focus on the terms with 10 as their factors.