One way to approach this is to write the division as a quotient and a remainder, like this: $109 = qd + 4$, where $q$ is the quotient. Then, I can rearrange this equation to get $105 = qd$, which means that $d$ is a divisor of $105$. Now, I need to find all the two-digit divisors of $105$. I can use prime factorization to help me do this. I notice that $105 = 3 \times 5 \times 7$, which are all prime numbers. Therefore, any divisor of $105$ must be a product of some combination of these three factors. The possible products are $1, 3, 5, 7, 15, 21, 35, 105$. However, not all of these are two-digit numbers. Only $15, 21, 35$ are. These are the only possible values of $d$ that satisfy the given condition. To find their sum, I just add them up: $15 + 21 + 35 = 71$.


Let $d$ be a positive number such that when $109$ is divided by $d$, the remainder is $4.$ Compute the sum of all possible two-digit values of $d$.