A two-digit integer between 10 and 99, inclusive, is chosen at random. Each possible integer is equally likely to be chosen. What is the probability that its tens digit is a multiple of its units (ones) digit?
The number of two-digit integers in the range 10 to 99 inclusive is $99-10+1=90$.

We count the number of two-digit positive integers whose tens digit is a multiple of the units (ones) digit.

If the units digit is 0 , there are no possible tens digits.

If the units digit is 1 , the tens digit can be any digit from 1 to 9 . This gives 9 such numbers.

If the units digit is 2 , the tens digit can be $2,4,6,8$. This gives 4 such numbers.

If the units digit is 3 , the tens digit can be $3,6,9$. This gives 3 such numbers.

If the units digit is 4 , the tens digit can be 4,8 . This gives 2 such numbers.

If the units digit is any of $5,6,7,8,9$, the only possibility is that the tens digit equals the units digit. This gives 5 such numbers.

In total, there are $9+4+3+2+5=23$ two-digit positive integers whose tens digit is a multiple of the units digit, and so the probability that a randomly chosen two-digit integer has this property is $\frac{23}{90}$.