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

Given:

The five-digit number 725yz is divisible by 15.

Formula used:

A number is divisible by 15 if it is divisible by both 3 and 5.

1. For divisibility by 5, the last digit (z) must be 0 or 5.

2. For divisibility by 3, the sum of the digits must be divisible by 3.

Calculation:

Sum of the digits of 725yz = 7 + 2 + 5 + y + z = 14 + y + z

For the number to be divisible by 3, 14 + y + z must be divisible by 3.

Let's consider the possible values for z (0 or 5):

1. If z = 0:

⇒ 14 + y + 0 = 14 + y

For 14 + y to be divisible by 3, y must be 1 or 4 or 7 (since 14 % 3 = 2, we need y % 3 = 1).

Possible pairs (y, z) are (1, 0), (4, 0), (7, 0).

Product (y × z) = 1 × 0 = 0, 4 × 0 = 0, 7 × 0 = 0

2. If z = 5:

⇒ 14 + y + 5 = 19 + y

For 19 + y to be divisible by 3, y must be 2 or 5 or 8 (since 19 % 3 = 1, we need y % 3 = 2).

Possible pairs (y, z) are (2, 5), (5, 5), (8, 5).

Product (y × z) = 2 × 5 = 10, 5 × 5 = 25, 8 × 5 = 40

∴ The maximum possible value of the product of y and z is 40.

∴ The correct answer is option (4).