The smallest number which when diminished by $7$, is divisible $12$, $16$, $18$, $21$ and $28$ is:

The correct option is B $1015$

First find the smallest number divisible by $12,16,18,21$ and $28$.

It is the LCM of these numbers.

$12=2\times 2\times 3=2^2\times 3$

$16=2\times 2\times 2\times 2=2^4$

$18=2\times 3\times 3=2\times 3^2$

$21=3\times 7$

$28=2\times 2\times 7=2^2\times 7$

$LCM=2^4\times 3^2\times 7=1008$

Required number $=$ (L.C.M. of $12$, $16$, $18$, $21$, $28$) $+7$

$=1008+7$
$=1015$

So the required number is $1015$.

If you diminish by $7,$ it will be divisible by all the given numbers.