Let x be the least number which when divided by 28, 40 and 48 leaves remainder 4 in each case and xdivisible by 13, What is the sum of the digits of x?

Correct option is B

Given :

x leaves remainder 4 when divided by 28, 40, 48

x is divisible by 13

Find the sum of digits of the least x

Solution:

Since the remainder is 4 in all cases

x - 4 is exactly divisible by 28,40,48

​$$\text{LCM}=2^4\times3\times5\times7=1680$$​​

So,

x=1680n+4

Now check divisibility by 13 (simple division):

1680 + 4 = 1684 (not divisible by 13)

​$$2\times1680+4=336$$​4 (not divisible)

3$$\times1$$​680+4=5044 (divisible by 13)

Hence,

x=5044

Sum of digits:

5+0+4+4=13