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

We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occurCase I : When n = 3qIn this case, we have,n = 3q, which is divisible by 3n = 3q= adding 2 on both sidesn + 2 = 3q + 2n + 2 leaves a remainder 2 when divided by 3Therefore, n + 2 is not divisible by 3n = 3qn + 4 = 3q + 4 = 3(q + 1) + 1n + 4 leaves a remainder 1 when divided by 3n + 4 is not divisible by 3Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3Case II : When n = 3q + 1In this case, we haven = 3q +1n leaves a reaminder 1 when divided by 3n is not divisible by 3n = 3q + 1n + 2 = (3q + 1) + 2 = 3(q + 1)n + 2 is divisible by 3n = 3q + 1n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2n + 4 leaves a remainder 2 when divided by 3n + 4 is not divisible by 3Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3Case III : When n = 3q + 2In this case, we haven = 3q + 2n leaves remainder 2 when divided by 3n is not divisible by 3n = 3q + 2n + 2 = 3q + 2 + 2 = 3(q + 1) + 1n + 2 leaves remainder 1 when divided by 3n + 2 is not divsible by 3n = 3q + 2n + 4 = 3q + 2 + 4 = 3(q + 2)n + 4 is divisible by 3Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3 . Thanks I HOPE YOU UNDERSTAND?
We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occurCase I :\xa0When n = 3qIn this case, we have,n = 3q, which is divisible by 3n = 3q= adding 2 on both sidesn + 2 = 3q + 2n + 2 leaves a remainder 2 when divided by 3Therefore, n + 2 is not divisible by 3n = 3qn + 4 = 3q + 4 = 3(q + 1) + 1n + 4 leaves a remainder 1 when divided by 3n + 4 is not divisible by 3Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3Case II :\xa0When n = 3q + 1In this case, we haven = 3q +1n leaves a reaminder 1 when divided by 3n is not divisible by 3n = 3q + 1n + 2 = (3q + 1) + 2 = 3(q + 1)n + 2 is divisible by 3n = 3q + 1n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2n + 4 leaves a remainder 2 when divided by 3n + 4 is not divisible by 3Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3Case III :\xa0When n = 3q + 2In this case, we haven = 3q + 2n leaves remainder 2 when divided by 3n is not divisible by 3n = 3q + 2n + 2 = 3q + 2 + 2 = 3(q + 1) + 1n + 2 leaves remainder 1 when divided by 3n + 2 is not divsible by 3n = 3q + 2n + 4 = 3q + 2 + 4 = 3(q + 2)n + 4 is divisible by 3Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3 .\xa0