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N=1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21…., 999

Divisible by either 3 or 5 [ P(3 or 5) ] = Divisible by 3 [ P(3) ] + Divisible by 5[ P(5) ] – Divisible by both 3 & 5 [ P(3 & 5) ]

P(3) :   3, 9, 15,21,27,….,999

AP series :  999 = 3 + [n-1]6    ;   n= 167    

P(5) :   5,15,25,35,…,995

995 = 5 + [n-1]10  ;   n = 100

P(3&5) :  15,45,75,105,135,..,990

990 = 15 + [n-1]30;  n= 33.5 => 33

So,  P(3 or 5) = 167 + 100 – 33 = 234

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