Question

prove that one of any three consecutive positive integers must be divisible by 3.

Answer 1

Let three consecutive positive integers be n, =n + 1 and n + 2
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. 
If n = 3p, then n is divisible by 3. 
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. 
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. 

Answer 2

Answer: Let the first of our three

consecutive numbers is N, so that they are N, N+1, and N+2. Now, our

number N must leave a remainder of either 0, 1, or 2 when divided by

3, so it can be written as either 3k, 3k+1, or 3k+2 for some k (the

quotient). So we have three cases:

N = 3k, N+1 = 3k+1, N+2 = 3k+2 ==> only N is a multiple of 3

N = 3k+1, N+1 = 3k+2, N+2 = 3k+3 ==> only N+2 is a multiple of 3

N = 3k+2, N+1 = 3k+3, N+2 = 3k+4 ==> only N+1 is a multiple of 3

So there is always exactly one multiple of 3 among them.