In the number $64376759247$, each even digit is increased by 2 and each odd digit is decreased by 2. After this transformation, find the sum of the digits located at the odd positions in the new number formed.

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**Solution:**

1. Write down the original number with digit positions:

| Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|----------|---|---|---|---|---|---|---|---|---|----|----|
| Digit    | 6 | 4 | 3 | 7 | 6 | 7 | 5 | 9 | 2 | 4  | 7  |

2. Apply the transformation rules:
   - Even digits: add 2
   - Odd digits: subtract 2

Transform each digit:
- Position 1: 6 (even) → 6 + 2 = 8
- Position 2: 4 (even) → 4 + 2 = 6
- Position 3: 3 (odd) → 3 - 2 = 1
- Position 4: 7 (odd) → 7 - 2 = 5
- Position 5: 6 (even) → 6 + 2 = 8
- Position 6: 7 (odd) → 7 - 2 = 5
- Position 7: 5 (odd) → 5 - 2 = 3
- Position 8: 9 (odd) → 9 - 2 = 7
- Position 9: 2 (even) → 2 + 2 = 4
- Position 10: 4 (even) → 4 + 2 = 6
- Position 11: 7 (odd) → 7 - 2 = 5

3. The new number formed is:

$$8 \, 6 \, 1 \, 5 \, 8 \, 5 \, 3 \, 7 \, 4 \, 6 \, 5$$

4. Identify digits at odd positions (1, 3, 5, 7, 9, 11):

- Position 1: 8
- Position 3: 1
- Position 5: 8
- Position 7: 3
- Position 9: 4
- Position 11: 5

5. Calculate the sum:

$$8 + 1 + 8 + 3 + 4 + 5 = 29$$

**Answer:** $29$

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