Question

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

Answer 1

We applied Euclid Division algorithm on n and 3. a = bq +r  on putting a = n and b = 3 n = 3q +r  , 0<r<3 i.e n = 3q   -------- (1),n = 3q +1 --------- (2), n = 3q +2  -----------(3) n = 3q is divisible by 3 or n +2  = 3q +1+2 = 3q +3 also divisible by 3 or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3 Hence n, n+2 , n+4 are 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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