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

there could be 3 possibilities for all numbers 3q,3q+1,3q+2
for n
n=3q now it is divisible by 3
n=3q+1 now it is not it will leave a remainder 1
n=3q+2 now it will leave a remainder 3

for n+1
n+1=3q
n=3q-1 not divisible 3
n+1=3q+1
n=3q divisible by 3
n+1=3q+2
n=3q+1 not divisible by 3

for n+2
n+2= 3q
n=3q-2 not divisible by 3
n+2=3q+1
n=3q-1 not divisible by 3
n+2=3q+2
n=3q divisible by 3
thus only one of them is divisible under a specific condition i.e if the number is 3q or 3q+1 or 3q+2

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.