After thinking about it, I realize that the problem can be solved without looking at every case, but the method is messy and takes a fair amount of work. It is worth doing only if the first set is fairly large and the second is fairly small. I will give a brief overview of what would have to be done. In the example that you give, first compute the size of the first set as 93 – 62 + 1 = 32. I am going to be looking at determining how many of these numbers are divisible by some number of the second set. To find out how many are not divisible, just subtract from 32.
To find out how many are divisible by 3, find 32/3 = 10. There may be one additional number that is divisible, which could be determined by use of modular arithmetic. You would do the same thing for 4, 5, 6 and 7. Add these values together.
Now you apply what is called the inclusion/ exclusion formula. The current value is too large because there has been double counting. To eliminate the double counting, look at each pair of numbers, say 3 and 4. To find how many numbers are divisible by both 3 and 4, find the least common multiple which is 3*4=12. Find out how many of the numbers of the first set are divisible by 12 and subtract from the previous.value. Do this for every pair of numbers.
Now you have to look at every set of 3 of the numbers, for example 3, 4 and 5. You want to add back how many are divisible by all three, which is just how many are divisible by the least common multiple, which in this case is 3*4*5=60. You would do this for every set of 3 whose product is not greater than 93.
Then you take numbers 4 at a time and subtract and then 5 at a time and add and so on. In the present example, it would not be necessary to go any further.