Prove that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.

Question

Philoid · Accepted Answer

Let a = n, b = n + 2 and c = n + 4

⇒ Order triplet is (a,b,c) = (n,n + 2,n + 4) where n is any positive integer i.e. 1, 2, 3....

At n = 1; (a,b,c) = (1,1 + 2,1 + 4) = (1, 3, 5)

At n = 2; (a,b,c) = (2,2 + 2,2 + 4) = (2, 4, 6)

At n = 3; (a,b,c) = (3,3 + 2,3 + 4) = (3, 5,7)

At n = 4; (a,b,c) = (4,4 + 2,4 + 4) = (4, 6, 8)

At n = 5; (a,b,c) = (5,5 + 2,5 + 4) = (5, 7,9)

At n = 6; (a,b,c) = (6,6 + 2,6 + 4) = (6, 8,10)

At n = 7; (a,b,c) = (7,7 + 2,7 + 4) = (7, 9,11)

At n = 8; (a,b,c) = (8,8 + 2,8 + 4) = (8, 10,12)

We observe that each triplet consist of one and only one number which is multiple of 3 i.e.

divisible by 3.

Hence, one and only one out of n. (n + 2) and (n + 4) is divisible by 3. where, n is any positive

integer.