Question

prove that one of every three consecutive positive integers s divisible by 3

Answer 1

n is any positive integer
b=3
By lemma
n=3q+r 0≤r<3
r=0 n=3q
r=1 n=3q+1
r=2 n=3q+2

n=3q n=3q+1 n=3q+2
<n=3q> n=3q+1 n=3q+2
n+2=3q+2 <n+2=3q+3> n+2=3q+4
n+4=3q+4 n+4=3q+5 <n+4=3q+6>

so one and only one out of n, n+2, n+4 is divisible by 3 i.e. three consecutive positive integers are divisible by 3.

Answer 2

Step-by-step explanation:

Let 3 consecutive positive integers be n, n + 1 and n + 2 .

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

n = 3p or 3p+1 or 3p+2, where p is some integer

If n = 3p = 3(p) , then n is divisible by 3

If n = 3p + 1, then n + 2 = 3p +1 + 2 = 3 p + 3 = 3 ( p + 1 ) is divisible by 3

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3

Hence it is solved.