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.

Answer 1

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Answer 2

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.

Answer :

$\sf\pink { Explanation}$

• let n be any positive integer.

Applying euclids division lemma to n and 3 .

$\sf\therefore{ n = 3q + r , 0 < or = r < 3 }$

Hence any positive integer n is of the form 3q, 3q + 1 , 3q + 2 .

Case - 1

If n = 3q then ,

n = 3q

It is divisible by 3

n + 2 = 3q + 2

It is not divisible by 3

n + 4 = 3q + 4

n + 4 = 3q + 3 + 1

n + 4 = 3 ( q + 1 ) + 1

n + 4 = 3m + 1

It is not divisible by 3 .

Case - 2

If n = 3q + 1 then ,

n = 3q + 1

It is not divisible by 3

n + 2 = 3q + 1 + 2

n + 2 = 3q + 3

n + 2 = 3 ( q + 1 )

n + 2 = 3m

It is divisible by 3

n + 4 = 3q + 4 + 1

n + 4 = 3q + 5

n + 4 = 3q + 3 + 2

n + 4 = 3 ( q + 1 ) + 2

n + 4 = 3m + 2

It is not divisible by 3 .

Case - 3

If n = 3q + 2 then ,

n = 3q + 2

It is not divisible by 3

n + 2 = 3q + 2 + 2

n + 2 = 3q + 4

n + 2 = 3q + 3 + 1

n + 2 = 3 ( q + 1 ) + 1

n + 2 = 3m + 1

It is not divisible by 3

n + 4 = 3q + 4 + 2

n + 4 = 3q + 6

n + 4 = 3 ( q + 2 )

n + 4 = 3m

It is divisible by 3 .