In the number $248375975$, how many pairs of digits have the same number of digits between them as in the original number sequence? Consider both forward and backward directions.

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**Solution:**

1. The given number is $248375975$.

2. We need to find pairs of digits that appear more than once with the same number of digits between them, considering both forward and backward directions.

3. Let's analyze the pairs and the gaps between their occurrences:

- Digit '2' appears only once, so no pair.

- Digit '4' appears once.

- Digit '8' appears once.

- Digit '3' appears twice: positions 4 and 7 (indexing from 1).
     - Between these two '3's, there are $7 - 4 - 1 = 2$ digits.

- Digit '7' appears twice: positions 5 and 9.
     - Between these two '7's, there are $9 - 5 - 1 = 3$ digits.

- Digit '5' appears twice: positions 6 and 8.
     - Between these two '5's, there is $8 - 6 - 1 = 1$ digit.

4. Now, check if these pairs have the same number of digits between them elsewhere in the number:

- For '3': only one pair, so counts as one.

- For '7': only one pair.

- For '5': only one pair.

5. Next, check pairs of different digits that appear with the same gap between them:

- Pair (4, 8) at positions 2 and 3: gap 0

- Pair (8, 3) at positions 3 and 4: gap 0

- Pair (3, 7) at positions 4 and 5: gap 0

- Pair (7, 5) at positions 5 and 6: gap 0

- Pair (5, 9) at positions 6 and 7: gap 0

- Pair (9, 7) at positions 7 and 8: gap 0

- Pair (7, 5) at positions 8 and 9: gap 0

6. These pairs with gap 0 appear multiple times, so counting all such pairs with the same gap, the total number of pairs having the same number of digits between them is more than four.

**Answer:** More than four

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