What is the sum of the digits of the least number which when divided by 12, 16 and 20 leaves the same remainder 6 in each case and it is divisible by 9?

First, find the Least Common Multiple (LCM) of the divisors 12, 16, and 20:
12=2
2
×3
16=2
4

20=2
2
×5
LCM(12,16,20)=2
4
×3×5=16×15=240
The required number must be of the form:

N=240k+6
where k is a positive integer such that N is completely divisible by 9.
Let us simplify 240k+6 with respect to modulo 9:

240k+6=(26×9+6)k+6=234k+6k+6
For N to be divisible by 9, the term (6k+6) must be divisible by 9.
Testing small values for k:

For k=1: 6(1)+6=12 (Not divisible by 9)

For k=2: 6(2)+6=18 (Divisible by 9)

So, the least value of k is 2.
Substitute k=2 back into the expression for N:

N=240×2+6=480+6=486
Now, find the sum of the digits of 486:

Sum of digits=4+8+6=18