If the number 732XY is divisible by 70, then find the minimum value of $\frac{x+y}{2}$.

1

If the number 732XY is divisible by 70

Then find the minimum value of $\frac{x+y}{2}$

If a number is divisible by 70 then it is also be divisible by 7 and 10

So, 732XY is divisible by 10 for the value of y = 0

and 732X0 is divisible by 7 for the value of x = 2

So, $\frac{2+0}{2}$ = 1