Question

1!+2!+3!..........+99! digit on unit position of this integer​

Answer 1

Answer: 3

Observe the unit position of 5! + 6! + 7! ..... + 99!

Note that the unit position of this expression is 0 , (because each of the factorials has a 5x2 multiplied with it , which gives 10 , and any no. multiplied by 10 has a 0 at its end.)

So we only have to deal with the unit digit of 1! + 2! + 3! + 4!

By manual checking , you will find that unit digit of :

1! + 2! + 3! + 4! is 3 .

⇒ 1 + 2 + 6 + 24

⇒ 33

So the answer is 3

Answer 2

Answer: 3

Step-by-step explanation:

1! + 2! + 3! + 4! + …… + 99!

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

From 5!, the numbers do not contribute to unit digit as the product contains at least one 2 and 5 resulting in 0 as the last digit in factorial Value.

•°• If we find the sum of these factorials, it will required unit digit

•°• Sum of these factorials = 1 + 2 + 6 + 24 + 0

= 33

Here, Unit digit = 3 in the resultant sum

•°• Required Unit digit = 3