The digit in the unit position of the integer 1! + 2! + 3! + ..... + 99! is

Question

Question

This question was previously asked in

RPSC 2nd Grade Mathematics (Held on 30th June 2017) Official Paper

Answer 1

Concept Used:

Factorial is defined as:

n! = n × (n - 1) × (n - 2) × … × 3 × 2 × 1

Calculation:

1! = 1 = 1 ⇒ Unit digit = 1

2! = 2 × 1 = 2 ⇒ Unit digit = 2

3! = 3 × 2 × 1 = 6 ⇒ Unit digit = 6

4! = 4 × 3 × 2 × 1 = 24 ⇒ Unit digit = 4

5! = 5 × 4 × 3 × 2 × 1 = 120 ⇒ Unit digit = 0

6! = 6 × 5! = 720 ⇒ Unit digit = 0

Therefore,

For the remaining number i.e. 7!, 8!, ....99!, the unit digit will be 0.

Hence, the required unit digit is the sum of unit digits

⇒ 1 + 2 + 6 + 4 + 0 + 0 + 0 + .....+ 0 = 13

Hence, the required unit digit is 3.

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