Question

prove that any of three consecutive positive integer is divisible by 3​

Answer 1

Answer:

I will prove.

proof-

if we take 123 as three consecutive positive integers it will divisible by 3.

123÷3=41.

Answer 2

HeRe Is Your Ans ⤵

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Let the first of our three consecutive numbers is N, so that they are N, N+1, and N+2

number N must leave a remainder of either 0, 1, or 2 when divided by

3 , so it can be written as either 3k, 3k+1, or 3k+2 for some k (the quotient)

So we have three cases :-

N = 3k , N+1 = 3k+1, N+2 = 3k+2 ==> only N is a multiple of 3

N = 3k+1, N+1 = 3k+2, N+2 = 3k+3 ==> only N+2 is a multiple of 3

N = 3k+2, N+1 = 3k+3, N+2 = 3k+4 ==> only N+1 is a multiple of 3

So there is always exactly one multiple of 3 among them

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$<marquee > hope it helps u$