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

Solution:
Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 3, we have
n = 3q + r …(i)
Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
n = 3q
So, n is divisible by 3.
n + 1 = 3q + 1
so, n + 1 is not divisible by 3.
n + 2 = 3q + 2
so, n + 2 is not divisible by 3.
Putting r = 1 in (i), we get
n = 3q + 1
so, n is not divisible by 3.
n + 1 = 3q + 2
so, n + 1 is not divisible by 3.
n + 2 = 3q + 3 = 3(q + 1)
so, n + 2 is divisible by 3.
Putting r = 2 in (i), we get
n = 3q + 2
so, n is not divisible by 3.
n + 1 = 3q + 3 = 3(q + 1)
so, n + 1 is divisible by 3.
n + 2 = 3q + 4
so, n + 2 is not divisible by 3.
Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.

Let 3 consecutive positive integers be n, n+1 and n+2 

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2. : 

Therefore:    

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 

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

Let 3 consecutive positive integers be p, p + 1 and p + 2

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

p = 3q or 3q + 1 or 3q + 2, where q is some integer

If p = 3q, then n is divisible by 3

If p = 3q + 1, then n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3

If p = 3q + 2, then n + 1 = 3q + 2 + 1 = 3q + 3 = 3(q + 1) is divisible by 3

Thus, we can state that one of the numbers among p, p + 1 and p + 2 is always divisible by 3