Understanding the Divisibility Rule for 15

A number is divisible by 15 if it is divisible by both 3 and 5. We need to apply these rules to the five-digit number 725yz.

Applying Divisibility by 5

For the number 725yz to be divisible by 5, its last digit, z, must be either 0 or 5.

Applying Divisibility by 3

For the number 725yz to be divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits is:

$$7 + 2 + 5 + y + z = 14 + y + z$$

This sum, $14 + y + z$, must be a multiple of 3.

Case Analysis for Maximum Product y * z

We will examine the two possible values for z (0 and 5) separately. For each value of z, we find the possible values of y that satisfy the divisibility by 3 rule. Then, we calculate the product $y \times z$ for each valid pair and identify the maximum value.

Case 1: z = 0

If $z = 0$, the sum of the digits becomes $14 + y + 0 = 14 + y$. We need $14 + y$ to be divisible by 3.

Since y must be a digit from 0 to 9:

• If $y=1$, the sum is $14 + 1 = 15$, which is divisible by 3. The number formed is 72510. The product $y \times z = 1 \times 0 = 0$.
• If $y=4$, the sum is $14 + 4 = 18$, which is divisible by 3. The number formed is 72540. The product $y \times z = 4 \times 0 = 0$.
• If $y=7$, the sum is $14 + 7 = 21$, which is divisible by 3. The number formed is 72570. The product $y \times z = 7 \times 0 = 0$.

In this case (when $z=0$), the maximum possible product $y \times z$ is 0.

Case 2: z = 5

If $z = 5$, the sum of the digits becomes $14 + y + 5 = 19 + y$. We need $19 + y$ to be divisible by 3.

Since y must be a digit from 0 to 9:

• If $y=2$, the sum is $19 + 2 = 21$, which is divisible by 3. The number formed is 72525. The product $y \times z = 2 \times 5 = 10$.
• If $y=5$, the sum is $19 + 5 = 24$, which is divisible by 3. The number formed is 72555. The product $y \times z = 5 \times 5 = 25$.
• If $y=8$, the sum is $19 + 8 = 27$, which is divisible by 3. The number formed is 72585. The product $y \times z = 8 \times 5 = 40$.

In this case (when $z=5$), the maximum possible product $y \times z$ is 40.

Determining the Overall Maximum Product

By comparing the maximum products obtained from both cases (Case 1: 0, Case 2: 40), we find that the overall maximum possible value for the product of y and z is 40.

Summary of Possibilities

The maximum product $y \times z$ across all valid possibilities is 40.