 # 11. What is the sum of digits of the leastmultiple of 13, which when divided by 6,8 and 12, leaves 5, 7 and 11 respectively,as the remainders?[CDS 2015 (11)(a) 5(b) 6(c) 7(d) 8​

The time for answering the question is over
391 cents dividing with 6. So I assume 7 remainder is for 8 and 5 is for 6.

First let's assume the answer is X

With these remainders, i.e, 5 when dividing with 6, 7 with 8 and 11 with 12, it is clear that X+1 will be a perfect multiple of 6,8 and 12

The LCM of 6,8,12 is 24. So, X+1 should be a multiple of 24

Rephrasing the problem,

We are looking for the least value of X where X is a multiple of 13 and X+1 is a multiple of 24

Which means X divided by 24 will give a remainder of 23.

Now let's look how each multiple of 13 gives a remainder when divided with 24

13%24 = 13

26%24 = 2

39%24 = 15

52%24 = 4

For every odd multiple, it is giving an odd remainder which is increasing by 2. Following this, we can find that

143%24 = 23

Checking

143%13 =0

143%6 = 5

143%8 = 7

143%12 = 11

that only all explanation

465  