Question

One of any three consecutive positive integers must be divisible by 3. Is it true if true enter 1 else0.

Answer 1

Let n be any positive number of the form $3q,3q+1,3q+2$ for any integer q
CASE:1 $n=3q$
$n=3q=3q+0$
As n leaves remainder 0 therefore n is divisible by 3
$n+2=3q+2$
as n+2 leaves remainder 2, therefore n+2 is not divisible by 3.
$n+4=3q+4=3q+3+1=3(q+1)+1$
As n+4 leaves remainder 1, therefore n+4 is not divisible by 3.
In Case1 when n=3q one and only one n out of $n,n+2,n+4$ is divisible by 3
CASE:2 $n=3q+1$
$n=3q+1$
As n leaves remainder 1 therefore n is not divisible by 3
$n+2=(3q+1)+2=3q+3=3(q+1)+0$
As n+2 leaves remainder 0 therefore n+2 is divisible by 3.
$n+4=(3q+1)+4=3q+3+2=3(q+1)+2$
As n+4 leaves remainder 2 therefore n+4 is not divisible by 3.
In Case2, when $n=3q+1$ , one and only on $n+2$ out of $n,n+2,n+4$ is divisible by 3
CASE:3 $n=3q+2$
$n=3q+2$
As n leaves remainder 2 therefore n is not divisible by 3.
$n+2=(3q+2)+2=3q+3+1=3(q+1)+1$
As n+2 leaves remainder 1 therefore n+2 is not divisible by 3.
$n+4=(3q+2)+4=3q+6=3(q+2)+0$
As n+4 leaves remainder 0, therefore n+4 is divisible by 3.
In case 3, when $n=3q+2$, one and only on $n+4$ out of $n,n+2,n+4$ is divisible by 3.
$\therefore$ from above case 1, 2,3 one and only one out of $n,n+2,n+4$ is divisible by 3.