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 q be quotient and r be the remainder.
On applying Euclid's algorithm, i.e. dividing n by 3, we have
      n = 3q + r       0 ≤ r < 3
⇒  n = 3q + r       r = 0, 1 or 2
⇒  n = 3q  or  n = (3q + 1) or n = (3q + 2)
Case 1​: If n = 3q, then n is divisible by 3.
Case 2: If n = (3q + 1), then (n + 2) = 3q + 3 = 3(3q + 1), which is clearly divisible by 3.
             In this case, (n + 2) is divisible by 3.
Case 3 : If n = (3q + 2), then (n + 4) = 3q + 6 = 3(q + 2), which is clearly divisible by 3.
              In this case, (n + 4) is divisible by 3.
Hence, one and only one out of n, (n + 1) and (n + 2) is divisible by 3.

Answer 2

Step-by-step explanation:

Question :-

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

▶ 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.