Question

Show that one and only out of N , N +2 , N + 4 is divisible by 3 . Where N is any positive integer

Answer 1

HOLA !!

----------------------
-----

Real numbers

--------------------------
------------

Given -- N is any positive integer

First case -- Let's take N = 3

So , 3 is divisible by 3

--------------------------------
------------

Second case -- Let's take N = 4

4 + 2 = 6

6 is divisible by 3

----------------------------------
-------------

Third case -- Let's take N = 2

2 + 4 = 6

6 is divisible by 3

---------------------------------------
--------------

Hence by all three cases we can say that any of the form N , N + 2 , N + 4 is divisible by 3 where the integer is positive

---------------------------------------
--------------

HOPE U UNDERSTAND

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.