Prove by induction that every integer can be represented in the form 3n, 3n+1, 3n+2. Never encountered such an induction question.
Base case N=1, this can be writen 3\times 0+1, so the proposition holds for N=1.
Suppose it true for N=k, then there is an n such that:
k=3n, or 3n+1 or 2n+3
Suppose the first of these is the case, then:
k+1=3n+1 and the proposition holds,
suppose the second is the case:
k+1=(3n+1)+1=3n+2, and the proposition holds,
suppose the third is the case:
k+1=(3n+2)+1=3(n+1), and the proposition holds.
Hence the proposition holds for k+1 whatever the case, and the induction goes through
RonL