Vtyjkyuk

I can't seem to wrap my head around this problem.How do I make sure that the digit 0 and digits from 7 to 9 are not included ? Provided- Repetition of digits is allowed.

To be a multiple of $3$, the sum of the three digits must be a multiple of $3$.
After choosing the first two digits freely, that specifies a congruence class
modulo $3$ for the last digit, and there are two possibilities within that range.
For instance, starting with $52$, the final digit must be $2$ or $5$, etc.

When fixing two of the digits i.e. hundreds and tens, we get exactly two numbers from $1-6$ which placed at ones digit will make the number divisible by $3$. You can prove this by observing that there are exactly $3$ numbers of the form $3n,3n+1,3n+2$ in $1-6$.

You are free to chose first two digits then you can have your last digit in two numbers.

So the number of numbers $=$ $6times6times2=72$

The trick is to note that for numbers that are divisible by 3, the sum of the digits in that number are also divisible by 3.

Start then by considering the three digit numbers that contain a 6. Since this is divisible by 3 already, the sum of the remaining two numbers must also be divisible by 3. The remaining 2 numbers can thus be (5,1), (5,4), (4,2), (6,6), (6,3), (3,3), (2,1) and so on.

Once you have the possible combinations of three digit numbers that add to 3, you then simply need to calculate the number of ways this three digit combination can be arranged.

Let be $A = 0, 1, 2 $ least residue system modulo 3.

Let be
$$
B_1 = (a, b, c) in Atimes Atimes A : a + b + c = 3
$$
$B_1$ has $frac5!3!2! - 3 = 7$ elements, then $2^3 cdot 7 = 56$ numbers with digits $1dots6$ such that the sun of these digits modulo 3 is 3.

At last,
$$
B_2 = (a, b, c) in Atimes Atimes A : a + b + c = 6 text or a + b + c = 0
$$
$B_2 = (0, 0, 0), (2, 2, 2) $, then, $16$ numbers with this property.

Hence, They are $72$ numbers.