The least number which when divided by 12, 16, 18, 30 leaves remainder 4 in each case but it is completely dividable by 7 is

$\therefore$  The L.C. M of 12, 16, 18 and 30 $=2\times 2\times 3\times 4\times 3\times 5=720$

Hence the number can be written in the term $(720\times K+4)$ which is divisible by 7.

If $K=1,number=(720\times 1)+4=724$ which is not divisible by ‘7’

If $K=2,number=(720\times 2)+4=1444$ which is not divisible by ‘7’

If $K=3,number=(720\times 3)+4=2164$ which is not divisible by ‘7’

If $K=4,number=(720\times 4)+4=2884$ which is divisible by ‘7’

$\therefore$ The required number is $=(720\times 4)+4=2884$