Question

No of positive integers less than or equal to 3600 which are prime to 3600

Answer 1

We basically need to find out the positive number less than or equal to 3600 not divisible by 3, 5 and 7, i.e. If we consider A,B, C respectively the events determining the total numbers divisible by 3,5,7

So, according to the question, we have to find numbers not divisible by 3,5 and 7, which is

= n (A’∩B’∩C’)

= n ((AꓴBꓴC)’) (By De Morgan’s Law)

= N- n (AUBUC) (Here N refers to total numbers in the sample space, i.e. 3600)

Consider n (AUBUC) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C)- n(A∩C) + n(A∩B∩C) (i)

Consider n (A) =3600/3=1200 (Total positive integers less than or equal to 3600 which are divisible by 3)

n (B) = 3600/5= 720 (Total positive integers less than or equal to 3600 which are divisible by 5)

n (C) = 3600/7=514 (Total positive integers less than or equal to 3600 which are divisible by 7)

n ( A∩B) =3600/15=420 (Total positive integers less than or equal to 3600 which are divisible by 3 and 5. Here we have taken the LCM of 3 and 5, 15)

n ( B∩ C) =3600/35=180 (Total positive integers less than or equal to 3600 which are divisible by 5 and 7. Here we have taken LCM of 5and 7, 35)

n(A∩C) = 3600/21 = 300 (Total positive integers less than or equal to 3600 which are divisible by 3 and 7. Here we have taken LCM of 3 and 7,21)

n(A∩B∩C)=3600/105=60 (Total positive integers less than or equal to 3600 which are divisible by 3,5 and 7. Here we have taken LCM of 3,5 and 7, 105)

Substituting all these values into equation (i) we get:

n(AUBUC)= 2100+1260+900-420-180-300+60=3420

Accordingly, N- n(AUBUC)= 3600-3420=180