If the six-digit number 15x1y2 is divisible by 44, then the minimum value of (x+y) is equal to:

For a number to be divisible by 44, it must be simultaneously divisible by both 4 and 11.
Divisibility by 4:
The last two digits of the number must be divisible by 4. The last two digits are y2.
Possible values for y such that y2 is divisible by 4 are 1,3,5,7,9.

Divisibility by 11:
The difference between the sum of digits at odd positions and the sum of digits at even positions must be either 0 or a multiple of 11.
Sum of digits at odd positions (from right, i.e., 2nd, 4th, 6th digits or vice versa):
Let's take positions from left to right:
Odd positions (1st, 3rd, 5th): 1+x+y
Even positions (2nd, 4th, 6th): 5+1+2=8

Difference = (1+x+y)−8=x+y−7

For divisibility by 11, x+y−7 must be 0 or a multiple of 11.
Case 1: x+y−7=0⟹x+y=7
Case 2: x+y−7=11⟹x+y=18

The minimum possible value of (x+y) is 7.
Let's check if x+y=7 works with any valid y:
If y=1, then x=6 (Valid, digits are single non-negative integers).
Thus, the minimum value of x+y is 7.