Question

use euclid division lemma to prove that one of any 3 consecutive positive integer must be divisibe by 3​

Answer 1

Step-by-step explanation:

29-May-2017 · 2 answers

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 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. So there is always exactly one multiple of 3 among them.

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