Question

show that one and only one number out of n, n+1 and n+2 is divided by 3 where n is any positive integer

Answer 1

sol) 
let "a" be any positive integer which is divisible by 3
i.e., according to euclid's division algorithm its 3p (or) 3p+1 (or) 3p+2

case 1

let n = 3p
n = 3p is divisible by 3
n + 2 = 3p +2 is not divisible by 3
n + 4 = 3p + 4 is not divisible by 3

case 2

let n = 3p+1
n = 3p+1 is not divisible by 3
n + 2 = (3p+1) +2 = 3p +3 is divisible by 3
n + 4 = (3p+1) + 4 = 3p + 5 is not divisible by 3

case 3 

let n = 3p + 2
n = 3p + 2 is not divisible by 3
n + 2 = (3p + 2) + 2 = 3p + 4 is not divisible by 3
n + 4 = (3p + 2) + 4 = 3p + 6 is divisible by 3

therefore one and only one out of n, n+1, n+2 is divisible by 3 in each case