Suppose that the number 735383629 is modified by subtracting 1 from each odd digit. If all the digits of the newly formed number are subsequently sorted in ascending order from left to right, what will be the sum of the digits occupying the 2nd, 4th, 6th, and 7th positions from the left?

The correct option is 16.

Step 1: Modify the original number
The given number is 735383629.
According to the rule, we subtract 1 from each odd digit and keep even digits unchanged:
• 1st digit: 7 (odd) ⇒
$7-1=6$
• 2nd digit: 3 (odd) ⇒
$3-1=2$
• 3rd digit: 5 (odd) ⇒
$5-1=4$
• 4th digit: 3 (odd) ⇒
$3-1=2$
• 5th digit: 8 (even) ⇒ 8
• 6th digit: 3 (odd) ⇒
$3-1=2$
• 7th digit: 6 (even) ⇒ 6
• 8th digit: 2 (even) ⇒ 2
• 9th digit: 9 (odd) ⇒
$9-1=8$
The newly formed set of digits is: 6, 2, 4, 2, 8, 2, 6, 2, 8.

Step 2: Sort the newly formed digits in ascending order
Arranging all digits from smallest to largest gives:
2, 2, 2, 2, 4, 6, 6, 8, 8

Step 3: Find the required digits and calculate their sum
Identifying the positions from left to right:
• 1st position: 2
• 2nd position: 2
• 3rd position: 2
• 4th position: 2
• 5th position: 4
• 6th position: 6
• 7th position: 6
• 8th position: 8
• 9th position: 8

Summing the digits at the 2nd, 4th, 6th, and 7th positions:
$\text{Sum}=2+2+6+6=16$
Thus, the sum of the specified digits is 16.