Question 1210205
Let a 7-digit sequence be represented by $d_1 d_2 d_3 d_4 d_5 d_6 d_7$, where each $d_i \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
We are looking for the number of sequences where a digit appears at least 6 times. This can happen in two cases:
Case 1: A digit appears exactly 6 times.
Case 2: A digit appears exactly 7 times.

Case 1: A digit appears exactly 6 times.
First, choose which digit appears 6 times. There are 10 choices (0 to 9).
Next, choose the positions for these 6 digits in the 7-digit sequence. There are $\binom{7}{6}$ ways to choose the 6 positions.
Finally, choose the digit for the remaining position. Since this digit cannot be the same as the digit that appears 6 times, there are 9 choices for this remaining digit.
So, the number of sequences where a digit appears exactly 6 times is $10 \times \binom{7}{6} \times 9 = 10 \times 7 \times 9 = 630$.

Case 2: A digit appears exactly 7 times.
First, choose which digit appears 7 times. There are 10 choices (0 to 9).
Next, all 7 positions in the sequence must be this chosen digit. There is only $\binom{7}{7} = 1$ way to place these digits.
So, the number of sequences where a digit appears exactly 7 times is $10 \times \binom{7}{7} = 10 \times 1 = 10$.

The total number of 7-digit sequences where a digit appears at least 6 times is the sum of the numbers from Case 1 and Case 2.
Total number of sequences = (Number of sequences with exactly 6 repetitions) + (Number of sequences with exactly 7 repetitions)
Total number of sequences = 630 + 10 = 640.

Final Answer: The final answer is $\boxed{640}$