27 integers. I'd go with a sets approach here.
T: # of integers between 100 and 150 incl. = 51
A: # of integers between 100 and 150 incl. that are multiples of 3 = # of integers between 34 and 50 incl. (think multiples of 3 between 102 and 150) = 17
B: # of integers between 100 and 150 incl. that are multiples of 5 = # of integers between 20 and 30 incl. = 11
C: Intersection set. Need the number of integers between 100 and 150 incl. that are divisible by both 3 and 5. Essentially, multiples of 15. So that is the same as saying # of integers between 7 and 10 incl. (think multiples of 15 between 105 and 150) = 4
So, # of integers between 100 and 150 incl. that can be evenly divided by neither 3 nor 5 = T-(A+B)+C = 51-28+4=27. [You need to add back the 4 to avaoid double counting, because those 4 are already included in sets A and B).
