What is the sum of digits of the least multiple of 13, which when divided by 6, 8 and 12 leaves 5, 7 and 11 respectively as the remainders?

8

Given:

In each case when the required number is divided by 6, 8, 12 leaves remainders 5, 7, 11 respectively and the required number will be the least multiple of 13

⇒ When the difference between divisor and remainder will same in each case, the difference should be subtracted from LCM to get the required number.

6 – 5 = 1

8 – 7 = 1

12 – 11 = 1

∴ The LCM of 6, 8, 12 = 24

Now it is given that the required number should be a multiplier of 13

∴ The required number should be (24x – 1)

Now, we can write (24x – 1) as (13x + 11x – 1), where 117x is exactly divisible by 13

∴ For x = 6, the remainder of (11x – 1) will be zero and is completely divided by 13

∴ So, the number is (24 × 6 – 1) = 143

The sum of digits of 143 = 8