Question

Show that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer

Answer 1

Let the positive integer in the form 3q, 3q +1, 3q +2

Case 1 n = 3q

n= 3q is divisible by 3

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

n+4 = 3q + 4 is not divisible by 3

Case 2 n= 3q + 1

n= 3q + 1 is not divisible

n + 2 = 3q + 1 + 2 = 3q + 3 is divisible by 3

n+ 4 = 3q + 4 not divisible

Case 3 n= 3q +2

n= 3q +2 is not divisible

n+2 = 3q +2 +2 = 3q +4 is not divisible

n+ 4 = 3q +2 + 4 = 3q +6 is divisible

Hence in each case one and only one out of n, n+2, n+4 is divisible by 3

Answer 2

Step-by-step explanation:

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

where r = 0, 1, 2,. and q is the quotient.

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

case I: if n =3q

n = 3q = 3(q) 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.

Hence, it is solved

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