17. Prove that one of every three consecutive positive integers is divisible by 3.
Answers
Let three consecutive positive integers be n, =n + 1 and n + 2
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 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
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
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.