What is the smallest number which, when divided by 15, 17 and 19, leaves remainder 4, 6 and 8, respectively?

Correct option is D

Given:

Given numbers are 15, 17 and 19

Formula Used:

LCM of prime numbers = Product of numbers

Solution:

Difference between divisor and remainder:

15 - 4 = 11

17 - 6 = 11

19 - 8 = 11

Here the difference between divisors and remainders is same in all three cases.

Now LCM of 15, 17 and 19 is $$15 \times 17 \times 19 = 4845$$​​

The smallest number giving remainder 4,6 and 8 is 4845 - 11 = 4834