Question 108

A number x is divisible by 7. When this number is divided by 8, 12 and 16, it leaves a remainder 3 in each case. The least value of x is:

Solution

The number x is divisible by 7, => $$x=7k$$

Now L.C.M. (8,12,16) = 48

Thus, the least number which is divided by 8, 12 and 16 and leaves a remainder 3 in each case = $$48n+3$$

Now, $$f(n)=(48n+3)$$ should be divisible by 7.

By putting $$n=1,2,3,...$$

$$f(1)=48(1)+3=51$$

$$f(2)=48(2)+3=99$$

$$f(3)=48(1)+3=147$$ which is divisible by 7.

=> Ans - (C)

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