A positive integer $n$ between 10 and 99 , inclusive, is chosen at random. If every such integer is equally likely to be chosen, what is the probability that the sum of the digits of $n$ is a multiple of 7 ?
There are $99-10+1=90$ positive integers between 10 and 99 , inclusive.

If a positive integer is between 10 and 99 , the minimum possible sum of its digits is $1+0=1$ and the maximum possible sum of its digits is $9+9=18$.

The multiples of 7 in the range 1 to 18 are 7 and 14 .

This means that we want to determine the number of positive integers between 10 and 99 , inclusive, with sum of digits equal to 7 or 14 .

The integers with sum of digits equal to 7 are $16,25,34,43,52,61,70$, of which there are 7 . The integers with sum of digits equal to 14 are $59,68,77,86,95$, of which there are 5 .

Thus, there are $7+5=12$ integers in the desired range with sum of digits divisible by 7 , and so the probability of choosing one of these integers at random is $\frac{12}{90}=\frac{2}{15}$.

ANSWER: $\frac{2}{15}$