Prove that one of every three consecutive positive integers is divisible by 3.

Explanation

Any three consecutive positive integers must be of the form n, (n+1) and (n+2), where n is any natural number i.e. n=1,2,3…
Let, a=n,b=n+1 and c=n+2
Order triplet is (a,b,c) (n,n+1,n+2), where n=1,2,3… (i)
`"At n=1 (a,b,c)=(1,1+1,1+2)=(1,2,3)"`
`"At n=2, (a,b,c)=(1,2+1,2+2)=(2,3,4)"`
`"At n=3, (a,b,c)=(3,3+1,3+2)=(3,4,5)"`
`"At n=4, (a,b,c)=(4,4+1,4+2)=(4,5,6)"`
`"At n=5, (a,b,c)=(5,5+1,5+2)=(5,6,7)"`
`"At n=6, (a,b,c)=(6,6+1,6+2)=(6,7,8)"`
`"At n=7, (a,b,c)=(7,7+1,7+2)=(7,8,9)"`
`"At n=8, (a,b,c)=(8,8+1,8+2)=(8,9,10)"`
We observe that each triplet consist of one and only one number which is multiple of 3 i.e. divisible by 3
Hence one of any three consecutive positive integers must be divisible by 3.