Question 1039608: Prove that one of every three consecutive positive numbers is divisible by 3
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! 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.
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Therefore:
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
Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3
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