Let A={n∈[100,700]∩N:n is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in A is [2024]
(2)
Multiples of 3 are 102,105,…,699
an=699=102+(n-1)·3
⇒597=3(n-1)⇒n=200
Now, multiples of 4 are 100,104,…,700
am=700=100+(m-1)·4
⇒m=151
Multiples of 3 and 4 are 108,120,…,696
ap=696=108+(p-1)·12
⇒p=50
n(3∪4)=n(3)+n(4)-n(3∩4)
=n+m-p=200+151-50=301
∴The number of elements in A=601-301=300