If a 9-digit number 937X728Y6 is divisible by 72, then one of the possible values of X+Y is:

For a number to be divisible by 72, it must be divisible by both 8 and 9 (since 8 and 9 are coprime factors of 72).
Step 1: Test for divisibility by 8
A number is divisible by 8 if its last three digits form a number divisible by 8. Here, the last three digits are 8Y6.
Let's find values of Y (0≤Y≤9) such that 8Y6 is divisible by 8:

If Y=1, 816/8=102 (Divisible)

If Y=5, 856/8=107 (Divisible)

If Y=9, 896/8=112 (Divisible)

So, possible values of Y are 1,5,9.

Step 2: Test for divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9.

Sum of digits=9+3+7+X+7+2+8+Y+6=42+X+Y
For (42+X+Y) to be divisible by 9, the nearest multiples of 9 greater than or equal to 42 are 45 and 54.

Case 1: If Y=1

42+X+1=43+X
For 43+X to be a multiple of 9 (i.e., 45), X=2.
Then, X+Y=2+1=3.

Case 2: If Y=5

42+X+5=47+X
For 47+X to be a multiple of 9 (i.e., 54), X=7.
Then, X+Y=7+5=12.

Case 3: If Y=9

42+X+9=51+X
For 51+X to be a multiple of 9 (i.e., 54), X=3.
Then, X+Y=3+9=12.

From the given options, 12 is available.