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

Let three consecutive integers are n, n + 1, n + 2 when n is a natural number
i.e., n = 1, 2, 3, 4, ….
for n = 1,         chosen numbers :(1, 2, 3) and 3 is divisible by 3
for n = 2,         chosen numbers :(2, 3,4) and 3 is divisible by 3
for n = 3,         chosen numbers :( 3, 4, 5) and 3 is divisible by 3
for n = 4,         chosen numbers :(4, 5, 6) and 6 is divisible by 3
for n = 5,         (chosen numbers :(5, 6, 7) and 6 is divisible by 3
Proof:

Case 1 : Let n is divisible by 3, it means n can be written as :
n = 3m,
Now, n +1 = 3m + 1 ; it gives remainder 1, when divided by 3
Now, n +2 = 3m + 2; it gives remainder 2, when divided by 3

Case 2: Let n + 1 is divisible by 3, it means n +1 can be written as :
n +1 = 3m,
Now, n = 3m – 1= 3(m – 1) + 2 ; it gives remainder 2, when divided by 3
Now, n +2 = 3m + 1; it gives remainder 1, when divided by 3

Case 3: Let n + 2 is divisible by 3, it means n +2 can be written as :
n +2 = 3m,
Now, n = 3m – 2= 3(m – 1) + 1 ; it gives remainder 1, when divided by 3
Now, n +1 = 3m – 1=3(m – 1) + 2  ; it gives remainder 2, when divided by 3