Find the no of positive integers less or equal to 3600 which are prime to 3600
Answers
Solution: 3600=(2^4)*(3^2)*(5^2).
So ,phi(3600)=(2^4-2^3)(3^2-3)(5^2-5)=8*6*20=960 is the correct and.
Answer:
No of Positive integers less or equal to 3600 which are prime to 3600 is 960.
Step-by-step explanation:
Prime Factorization of 3600 =
So, We basically need to find out the positive number less than or equal to 6300 not divisible by 2, 3 and 5, i.e. If we consider A,B, C respectively the events determining the total numbers divisible by 2,3,5
So, according to the question, we have to find numbers not divisible by 2,3 and 5, which is
= n (A’∩B’∩C’)
= n ((A∪B∪ C)’) (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) ......(1)
Consider n (A) =3600/2=1800 (Total positive integers less than or equal to 3600 which are divisible by 2)
n (B) = 3600/3=1200 (Total positive integers less than or equal to 3600 which are divisible by 3)
n (C) = 3600/5=720 (Total positive integers less than or equal to 3600 which are divisible by 5)
n ( A∩B) =3600/6=600 (Total positive integers less than or equal to 3600 which are divisible by 2 and 3. Here we have taken the LCM of 2 and 3 is 6)
n ( B∩ C) =3600/15=240 (Total positive integers less than or equal to 3600 which are divisible by 3 and 5. Here we have taken LCM of 3 and 5 is 15)
n(A∩C) = 3600/10 = 360 (Total positive integers less than or equal to 3600 which are divisible by 2 and 5. Here we have taken LCM of 2 and 5 is 10)
n(A∩B∩C)=3600/30=120 (Total positive integers less than or equal to 3600 which are divisible by 2,3 and 5. Here we have taken LCM of 2,3 and 5 is 30)
Substituting all these values into equation (i) we get:
n(AUBUC)= 1800 + 1200 + 720 - 600 -240 - 360 +120 = 2640
Accordingly, N- n(AUBUC)= 3600-2640 = 960
Therefore, No of Positive integers less or equal to 3600 which are prime to 3600 is 960.