An eight-digit number, 6x2y824 is divisible by 72. Find the maximum possible value of x + y.

Given number is 6x2y824. For a number to be divisible by 72, then it should be divisible by both 8 and 9. So, we will check the divisibility rule of 8 and 9. For the given number to be divisible by 8, the last three digits of the number should be divisible by 8. Last three digits of the given number are 824. So, the possible values of y are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (since 824 is already divisible by 8). For the number to be divisible by 9, the sum digits of the given number should be a multiple of 9. So, sum of the digits of the given number
= 6 + x + 2 + y + 8 + 2 + 4 = 22 + x + y So, next multiples of 9 after 22 are 27 and 36. So, possible values of (x + y) are 5 and 14. When, x + y = 14, take y = 9 and x = 5. So, this case is valid and sufficient to conclude that maximum value of x + y = 14. Hence, option B.