The smallest number which when added to 10000 becomes divisible by 20, 24 and 30 is

Correct option is C

Given:
We need to find the smallest number that should be added to 10000 so that the result is divisible by 20, 24, and 30.

Concept Used:
Find the LCM (Least Common Multiple) of 20, 24, and 30.
Then find the remainder when 10000 is divided by that LCM.
Then subtract the remainder from LCM to get the smallest number to be added.

Solution:
Step 1: Find LCM of 20, 24, and 30.

Prime factorization:
20 = 2² × 5
24 = 2³ × 3
30 = 2 × 3 × 5

LCM = 2³ × 3 × 5 = 120

Step 2: Divide 10000 by 120
10000 ÷ 120 = 83 remainder 40
So, 10000 leaves a remainder of 40 when divided by 120

Step 3: We need to add (120 − 40) = 80 to 10000

Ans. (c) 80