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

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CASE 3 When n=3q+2
In this case we have
n=3q+2
=>n leaves remainder 2 when divided by 3
=>n is not divisible by 3

Now,n=3q+2
=>n+2=3q+2+2=3(q+1)+1
=>n+2 leaves remainder 1 when divided by 3
=>n+2 is not divisible by 3

Again, n=3q+2
n+4=3q+2+4=3(q+2)
=>n+4 is divisible by 3

Thus,n+4 is divisible by 3 but n, n+2 are not divisible by 3



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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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