The five-digit number 725yz is divisible by 15. What is the maximum possible value of the product of y and z?

For a number to be divisible by 15, it must be divisible by both 3 and 5.
Divisibility by 5: The last digit z must be either 0 or 5.
To maximize the product y×z, we choose z=5 (since z=0 would make the product 0).

Divisibility by 3: The sum of digits must be a multiple of 3.

Sum of digits=7+2+5+y+5=19+y
For (19+y) to be a multiple of 3, the possible values of single-digit y are:

If 19+y=21⟹y=2

If 19+y=24⟹y=5

If 19+y=27⟹y=8

The maximum value of y is 8.

Maximum Product:

Maximum Product=y×z=8×5=40