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[heavy hound]

@lucid copper in case of first 4 digits, you considered only 8 choices for for 5th digit but there would be 9 choices (0-9, except for the consecutive digit). Also, for the 6th digit, there would be 10 choices (0-9).

[lucid copper]

You are correct, I made a mistake in my calculations. Let's correct the approach for 4 consecutive equal digits:

1. 4 consecutive equal digits:

• First 4 digits: 9 choices for the consecutive digits (1-9) × 9 choices for the 5th digit (0-9, except for the consecutive digit) × 10 choices for the 6th digit (0-9) = 9 × 9 × 10
• Middle 4 digits: 9 choices for the 1st digit (1-9) × 9 choices for the consecutive digits (0-9, except the 1st digit) × 9 choices for the 6th digit (0-9, except the consecutive digit) = 9 × 9 × 9
• Last 4 digits: 9 choices for the 1st digit (1-9) × 10 choices for the 2nd digit (0-9) × 9 choices for the remaining consecutive digits (0-9, except the 2nd digit) = 9 × 10 × 9
  Total for 4 consecutive equal digits: (9 × 9 × 10) + (9 × 9 × 9) + (9 × 10 × 9) = 90 × 9 + 81 × 9 × 2 = 810 × 2 = 1620 numbers

2. 5 consecutive equal digits:
   This calculation remains the same as before.
   Total for 5 consecutive equal digits: 2 × (9 × 9) = 162 numbers

3. 6 consecutive equal digits:
   This calculation remains the same as before.
   Total for 6 consecutive equal digits: 9 numbers

Now, adding up the numbers for all three scenarios: 1620 + 162 + 9 = 1791

There are 1791 six-digit numbers with more than 3 equal and consecutive digits. I apologize for the errors in my previous responses, and I appreciate your attention to detail.

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[heavy hound]

@lucid copper you did a mistake in summing up the total cases for 4 consecutive equal digits

[lucid copper]

Apologies for that mistake. Let me correct the totals sum for the 4 consecutive equal digits:

First 4 digits: 9 × 9 × 10 = 810
Middle 4 digits: 9 × 9 × 9 = 729
Last 4 digits: 9 × 10 × 9 = 810

Total for 4 consecutive equal digits: 810 + 729 + 810 = 2349 numbers

Now, let's sum the totals for all three scenarios:

Total for 5 consecutive equal digits: 162 numbers
Total for 6 consecutive equal digits: 9 numbers

Final sum: 2349 (4 consecutive equal digits) + 162 (5 consecutive equal digits) + 9 (6 consecutive equal digits) = 2520

So, there are 2520 six-digit numbers with more than 3 equal and consecutive digits. Apologies for the confusion earlier, and thank you for pointing it out.

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[heavy hound]

@lucid copper in a similar way, can you do it for 7 digit numbers?