If the 8-digit number 1a765b12 is to be divisible by 72, the least value of (2a + 3b)is:

a and b are digits of the number

$$=$$> $$0\le a\le9$$ and $$0\le b\le9$$

$$=$$> $$0\le a+b\le18$$ .................(1)

Given, the number 1a765b12 is divisible by 72

$$=$$>  The number is divisible by both 9 and 8

If a number is divisible by 9 then the sum of the digits of the number is divisible by 9

$$=$$>  1+a+7+6+5+b+1+2 is divisible by 9

$$=$$>  22+a+b is divisible by 9 ..........(2)

From (1) and (2), the possible values of a+b = 5,14

For the minimum value of 2a+3b,  value of a+b = 5 .......(3)

If a number is divisible by 8 then the last three digits of the number is divisible by 8

$$=$$>  b12 is divisible by 8

The possible values of b are 1,3,5,7,9

Substituting value of b in equation (3)

when b = 1, a = 5-1 = 4

when b = 3, a = 5-2 = 3

when b = 5, a = 5-5 = 0

when b is 7,9 the value of a is negative which not possible

The value of 2a+3b when a = 4 and b = 1 is 2(4)+3(1)=11

The value of 2a+3b when a = 3 and b = 3 is 2(3)+3(3)=15

The value of 2a+3b when a = 0 and b = 5 is 2(0)+3(5)=15

$$\therefore\ $$The minimum value of 2a+3b = 11

Hence, the correct answer is Option A