Lesson How many integer numbers in the range 1-300 are divisible by at least one of the integers 4, 6 and 15 ?

How many integer numbers in the range 1-300 are divisible by at least one of the integers 4, 6 and 15 ?

Problem 1

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.

Problem 2

We have a universal set U of 600 elements (integer numbers from 1 to 600 inclusive).


Of them, 600/4  = 150  elements are divisible by  4  (subset F, from the word Four);

         600/5  = 120  elements are divisible by  5  (subset V, from the word fiVe);

         600/6  = 100  elements are divisible by  6  (subset X, from the word siX).


Of them, we have these in-pair intersections

         600/(4*5)  = 30  elements divisible by 4 and by  5  (intersection (F and V) );

         600/(4*6) =  25  elements divisible by 4 and by  6  (intersection (F and X) );

         600/(5*6)  = 20  elements divisible by 5 and by  6  (intersection (V and X) ).


Of them, we have triple intersection

         600/(3*4*5) = 10  elements divisible by 4, 5 and 6  (intersection (F and V and X) ).



Having it, we can calculate the number of elements in the union of the three subsets (F U V U X).


Use the formula for the number of elements in the union of any 3 subsets (inclusion-exclusion principle)


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

                 =             substitute the obtained numbers from above                       = 

                 =  150 + 120 + 100  -    30      -     25       -     20     +   10 = 305.       


The rest  600 - 305 = 295  integer numbers from 1 to 600 inclusive are not divisible NEITHER by 4  or by 5  or by 6.


ANSWER.   There are 295 integer numbers from 1 and 600 (inclusive) that are NOT divisible  by 4  or by 5  or by 6.