We have a universal set U of 300 elements (integer numbers from 1 to 300).


Of them, 300/4  = 75  elements are divisible by  4  (subset F, from the word Four);

         300/6  = 50  elements are divisible by  6  (subset X, from the word siX);

         300/15 = 20  elements are divisible by 15  (subset N, from the word fifteeN).


Of them, we have these in-pair intersections

         300/(4*3)  = 25  elements divisible by 4 and by  6  (intersection (F and X) );

         300/(4*15) =  5  elements divisible by 4 and by 15  (intersection (F and N) );

         300/(6*5)  = 10  elements divisible by 6 and by 15  (intersection (X and N) ).


Of them, we have triple intersection

         300/(4*3*5) = 5  elements divisible by 4, 6 and 15  (intersection (F and X and N) ).



The problems asks about the number of elements in the union of the three subsets (F U X U N).


Use the formula for the number of elements in the union of any 3 subsets


    n(F U X U N) = n(F) + n(X) + n(N) - n(F and X) - n(F and N) - n(X and N) + n(F and X and N) = 

                 =             substitute the obtained numbers from above                       = 

                 =  75  +  50  +  20  -    25      -    5       -     10     + 5 = 110.              ANSWER


ANSWER.   There are 110 numbers between 1 and 300 (inclusive) that are divisible by at least one of three numbers 4, 6 and/or 15.