Question

Use euclid's division lemma to prove that one of any three consecutive positive integers must be divisible by 3

Answer 1

Answer:

Let 3 consecutive positive integers be p, p + 1 and p + 2

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

:

Therefore:

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

If p = 3q, then n is divisible by 3

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

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

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

Step-by-step explanation:

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