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

Checking the argument :- Only one of n, n+2 ,n+4 would be divisible by 3 when n is +ve integer.

Case 1 :-

n is divisible by 3 .

then n+2/3 = n/3 + 2/3 this shows a remainder 2 .

then n+4/3 =1+ n/3 + 1/3 this shows a remainder 1.

Case 2 :-

n+2 is divisible by 3 .

n/3 = n+2-2/3 = n+2/3 - 2/3 this leaves a remainder (3-2) = 1 .

n+4/3 =n+2+2/3 = n+2/3 + 2/3 leaves a remainder 2 .

Case 3 :-

n+4 is divisible by 3 .

n/3 = n+4-4/3 = (n+4)/3 -1 - 1/3 this leaves a remainder (3-1) = 2.

n+2/3 =n+4-2/3 = n+4/3 - 2/3 leaves a remainder (3-2)= 1 .

Hence proved that; one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer

Answer 2

Step-by-step explanation:

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 .

▶ 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.