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

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Here,

It is given that

n is an positive integer

So,

n = 3c + r

Here c is some integer and also 0 ≤ r <3

As per,

n = 3c

n = 3c + 1

n = 3c + 2

${\boxed{\sf\:{First\;Situation}}}$

Here

$\textbf{\underline{n = 3c ,\;n + 2 = 3c + 2 \;, n + 4 = 3c + 4}}$

We get that here only 3c is divisible by 3

${\boxed{\sf\:{Second\;Situation}}}$

Here

$\textbf{\underline{n = 3c + 1\;, n + 2 = 3c + 3 \;, n + 4 = 3c + 5}}$

We get that here,

n + 2 = 3c + 3

= 3(c + 1) it is divisible by 3

Next, n and n + 4 is not divisible by 3

${\boxed{\sf\:{Third\;Situation}}}$

Here,

n = 3c + 2 , n + 2 = 3c + 4 , n + 4 = 3c + 6

We get that

n + 4 = 3(c + 2) is divisible by 3

Therefore we get,

$\Large{\boxed{\sf\:{Only\;one\;out\;of\;n, n + 2 , n + 4\;is \;divisible\;by\;3}}}$

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 solvedv. mmmm

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