To find how many 3-digit numbers divisible by 3 can be formed from the digits $1, 2, 3, 4, 5, 6$ when repetition of digits is allowed, we need to use the rule of divisibility by 3.

Rule of Divisibility by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.

Step-by-Step Solution:

1. Total possible 3-digit numbers:

   • The number must be a 3-digit number, so we can choose any digit from $1, 2, 3, 4, 5, 6$ for each of the three positions (hundreds, tens, and ones).
   • There are 6 choices for each digit (since repetition is allowed), so the total number of 3-digit numbers is: $6 \times 6 \times 6 = 216$

2. Divisibility by 3:

   • We want to count how many of these 216 numbers are divisible by 3. For this, we need the sum of the digits $a + b + c$ (where $a$, $b$, and $c$ are the hundreds, tens, and ones digits) to be divisible by 3.

3. Possible sums of digits modulo 3: The digits we are working with (1, 2, 3, 4, 5, 6) have the following remainders when divided by 3:

   • $1 \equiv 1 \pmod{3}$
   • $2 \equiv 2 \pmod{3}$
   • $3 \equiv 0 \pmod{3}$
   • $4 \equiv 1 \pmod{3}$
   • $5 \equiv 2 \pmod{3}$
   • $6 \equiv 0 \pmod{3}$

   Therefore, the digits $1, 2, 4, 5$ contribute remainders of $1, 2, 1, 2$, and the digits $3, 6$ contribute remainders of $0$.

4. Counting valid numbers:

   • For each digit of the 3-digit number, we have 6 choices, but we only care about those combinations where the sum $a + b + c$ is divisible by 3.

   This problem is equivalent to counting how many valid combinations of remainders (modulo 3) sum to 0 modulo 3. Since there are exactly 2 choices each for each position, the pattern simplifies as the combinations all can be solved