show that one and only out of n , n+2 , n+4 is divisible by 3. where n is a positive integer

by Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,.........(a-1). and q is the quotient.

put a = 3: b = 3q+r and r = 0,1,2.

thus any number is in the form of 3q , 3q+1 or 3q+2.

case I: if n =3q

n is divisible by 3,

n+2 = 3q+2 is not divisible by 3.

n+4 = 3q+4 = 3(q+1)+1 is not divisible by 3.

case II: if n =3q+1

n = 3q+1 is not divisible by 3.

n+2 = 3q+1+2=3q+3 = 3(q+1) is divisible by 3.

n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2 is not divisible by 3.

case III: if n = 3q+2

n =3q+2 is not divisible by 3.

n+2 = 3q+2+2 =3q+4 = 3(q+1)+1 is not divisible by 3.

n+4 = 3q+2+4 = 3q+6 = 3(q+2) is divisible by 3.

thus one and only one out of n , n+2, n+4 is divisible by 3.

hope this helps you.

cheers!!

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