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Prove that one of any three consecutive positive integer is divisible by 3 |
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Answer» 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. |
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