Prove by induction that for all natural numbers, n, either 3|n or 3|n+1 or 3|n+2? That is prove that for all natural numbers, n, either 3 is a factor of n or n+1 or n+2
 A: Base case: n = 1.
Then 3 divides 3, which is n+2.
Inductive hypothesis: 
if the proposition is true for n: Then proof by cases
1). If n is divisible by 3, this implies that (n+1) + 2 is also divisible by 3.
2). If n +1 is divisible by 3, then (n+1) is also divisible by 3.
3). If n+2 is divisible by 3, then (n+1) + 1 is also divisible by 3.
Thus, for all natural numbers ,n, the 3 divides either n, n+1, or n+2.
A: (Base case: n=1): 3 is a factor of 3 = n+2.  
Inductive step.  
Assume that 3 is a factor of k.    
Then there exists an integer p such that $k = 3p.$    
This means that $(k+1)+2 =k+3 = 3(p+1)$ and so 3 is a factor of (k+1) + 2.  
The result follows by induction.   
A: Here is another hint for a different way of doing the induction step.
For $n$ the three numbers are $n, n+1, n+2$
For $n+1$ the three numbers are $n+1, n+2, n+3$
Consider two cases - that one of the common numbers is divisible by $3$ (easy); or ... (you fill in the gap).
A: $P(n) = 3 | (n + 0) \lor 3 | (n + 1) \lor 3 | (n + 2) \tag {to prove}$
$P(0)$ is true trivially.
Now we must establish $P(n) \rightarrow P(n + 1)$
Assume: $3 | (n + 0) \lor 3 | (n + 1) \lor 3 | (n + 2) \tag {inductive hypothesis}$
Establish:  $3 | (n + 1) \lor 3 | (n + 2) \lor 3 | (n + 3) \tag {inductive result}$
Break the inductive assumption into cases:
if $3|(n + 0)$ then $3|(n + 3)$, inductive result follows
if $3|(n + 1)$ then $3|(n + 1)$, inductive result follows
if $3|(n + 2)$ then $3|(n + 2)$, inductive result follows
