1. Set up the Divisibility Rules:
For a number to be divisible by 6, it must be an even number (divisible by 2) and the sum of its digits must be a multiple of 3 (divisible by 3).
- To make it an even number, the last digit $$d_7$$ can only be chosen from the even digits in our pool: 2 or 6.
- To make it divisible by 3, the total sum $$S_7 = d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7$$ must be divisible by 3.
2. Group the allowed digits by their remainders modulo 3:
- Remainder 0: The digit 6 (exactly 1 choice)
- Remainder 1: The digit 7 (exactly 1 choice)
- Remainder 2: The digits 2 and 5 (exactly 2 choices)
3. Trace the counting step-by-step for the first 6 digits:
Let $$a_n$$, $$b_n$$, and $$c_n$$ denote the number of ways to form an $$n$$-digit number whose sum leaves a remainder of 0, 1, and 2 when divided by 3, respectively.
For the very first digit ($$n = 1$$):
- Remainder 0 ($$a_1$$): 1 choice (the digit 6)
- Remainder 1 ($$b_1$$): 1 choice (the digit 7)
- Remainder 2 ($$c_1$$): 2 choices (the digits 2 or 5)
When adding each subsequent digit, the new counts are determined by multiplying the previous counts by the number of digit choices available for each remainder:
$$a_{n+1} = a_n \times 1 + b_n \times 2 + c_n \times 1$$
$$b_{n+1} = a_n \times 1 + b_n \times 1 + c_n \times 2$$
$$c_{n+1} = a_n \times 2 + b_n \times 1 + c_n \times 1$$
Let us compute the values systematically for up to 6 digits ($$n = 6$$):
- For $$n = 2$$:
$$a_2 = 1(1) + 1(2) + 2(1) = 5$$
$$b_2 = 1(1) + 1(1) + 2(2) = 6$$
$$c_2 = 1(2) + 1(1) + 2(1) = 5$$
- For $$n = 3$$:
$$a_3 = 5(1) + 6(2) + 5(1) = 22$$
$$b_3 = 5(1) + 6(1) + 5(2) = 21$$
$$c_3 = 5(2) + 6(1) + 5(1) = 21$$
- For $$n = 4$$:
$$a_4 = 22(1) + 21(2) + 21(1) = 85$$
$$b_4 = 22(1) + 21(1) + 21(2) = 85$$
$$c_4 = 22(2) + 21(1) + 21(1) = 86$$
- For $$n = 5$$:
$$a_5 = 85(1) + 85(2) + 86(1) = 341$$
$$b_5 = 85(1) + 85(1) + 86(2) = 342$$
$$c_5 = 85(2) + 85(1) + 86(1) = 341$$
- For $$n = 6$$:
$$a_6 = 341(1) + 342(2) + 341(1) = 1366$$
$$b_6 = 341(1) + 342(1) + 341(2) = 1365$$
$$c_6 = 341(2) + 342(1) + 341(1) = 1365$$
Thus, for the first 6 digits:
- There are 1366 ways to get a sum with a remainder of 0.
- There are 1365 ways to get a sum with a remainder of 1.
4. Apply the final constraint for the 7th digit ($$d_7$$):
We have two choices for the final even digit to complete the 7-digit number:
- Case 1: The final digit is 6 (remainder 0).
For the total sum to be divisible by 3, the sum of the first 6 digits must also leave a remainder of 0.
$$\text{Ways} = a_6 \times 1 = 1366 \times 1 = 1366$$
- Case 2: The final digit is 2 (remainder 2).
For the total sum to be divisible by 3, the sum of the first 6 digits must leave a remainder of 1 (since $$1 + 2 = 3$$, which is divisible by 3).
$$\text{Ways} = b_6 \times 1 = 1365 \times 1 = 1365$$
5. Sum the mutually exclusive cases:
$$\text{Total Numbers} = 1366 + 1365 = 2731$$
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