Video solution:

Click to load, then click again to play

Written solution:

For a number to be divisible by $15$ the number must be divisible by $3$ and by $5.$

First, to ensure the number is divisible by $5$ it must end in $0$ or in $5.$

Since the digits alternate between two distinct digits, we call the other digit $d.$

This would make our number either $d0d0d$ or $5d5d5.$ Since the last digit must be $0$ or $5$ and the first digit cannot be $0,$ the number must have the form $5d5d5.$

Thus, we know our number is $5d5d5$ for some digit $d.$

To ensure our number is also a multiple of $3$ the sum of the digits must be a multiple of $3.$

The sum of our digits is $5+d+5+d+5 = 15+2d.$ Since $15$ is a multiple of $3,$ all that is required is that $2d$ is a multiple of $3,$ so $d$ is a multiple of $3.$

This means $d$ can be $0,3,6$ or $9.$

Therefore, we have $4$ solutions.

Thus, the correct answer is B.