Question

show that one and only one out of n, n + 2 or n + 4 is divisible by 3 Where n is any positive integer​

Answer 1

Answer:

if n divisible by 3 then n = 3x

(n+2)/3 = (3x + 2)/3 = x + 2/3 = not divisible by 3

(n+4)/3 = (3x + 4)/3 = x + 4/3 = not divisible by 3

so n and n+2, and n and n+4 cannot be divisible by 3 at the same time

if n+2 divisible by 3 then n+2 = 3x and n = 3x -2

n/3 = (3x-2)/3 = x-2/3

(n+4)/3 = (3x+2)/3 = x+2/3

so n and n+2, and n+2 and n+4 cannot be divisible by 3 at the same time

Step-by-step explanation:

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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