We know that a number is divisible by 6 if it is:

Even

Divisible by 3 (for which the sum of digits should be divisible by 3)

Both of these facts can be checked by having a look at the prime factorization of 6, i.e., 2×3.

So we have to make numbers ensuring the above mentioned condition.

For a number to be even, the only simple condition is that the units place should be even. So we can have either 0, 2 or 4 at units place. That means the units place can have only 3 possible numbers.

Next up, we can have a case by case study of the remaining three numbers to see in what ways can we make the number a factor of 3.

If the units place is 0, then the required number should be in the form 3n or 3n + 0. Which means we have to pick numbers up from our list so that we end up getting the sum of those numbers as 3n. (n is any arbitrary non negative integer)

If it is 2, then we require the sum of the remaining numbers to be 3n-2 or 3k+1, for the sum to be divisible by 3.

If it is 4, then we require the sum of the remaining numbers to be 3n-4 or 3k+2 for the same.

But wait, haven't we covered all the possible forms in which a number could be? Any integer can be expressed in the form 3n, 3n+1 or 3n+2, all of which are obtained in the three cases above.

This means we can have any configuration of numbers and their sum would definitely land up in one of the three cases.

That means we can any of the 6 numbers in the thousands, hundreds and tens place.

Therefore, by the multiplication theorem, the total number of cases are: 6×6×6×3 = 648.

Find quatest 6 digit formed by using 2.05 with repetition