11. 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?
[CDS 2015 (11)
(a) 5
(b) 6
(c) 7
(d) 8
Answers
Answer:
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
Answer is 1+4+3 =8
Step-by-step explanation:
that only all explanation