SSC CHSL 15 Nov 2015 Shift 2 Paper

The number x is divisible by 7, ⇒ $\mathit{x}=7\mathit{k}$
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 = $48\mathit{n}+3$
Now, $\mathit{f}\left(\mathit{n}\right)=\left(48\mathit{n}+3\right)$ should be divisible by 7.
By putting $\mathit{n}=1,2,3,...$
$\mathit{f}\left(1\right)=48\left(1\right)+3=51$
$\mathit{f}\left(2\right)=48\left(2\right)+3=99$
$\mathit{f}\left(3\right)=48\left(1\right)+3=147$ which is divisible by 7.