Find the least number which when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7.

LCM of 16, 18 and 20 = 720
∴ Required number = 720k + 4
Where, k is a natural number to be divisible by 7, (720 k + 4) will be a multiple of 7.
Smallest value of k = 4
∴ Required number = 720 x 4 + 4 = 2884