Math, asked by mohammedfaizan258, 1 year ago

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prove that one of every three consecutive positive integers is divisible by 3.
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Answers

Answered by namitrustagi11
0

one of every three consecutive positve integers are divisible by three as there is gap of 3 no. between consecutive integers so thats why they r divisible by 3


Answered by Anonymous
0

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.


THANKS



#BeBrainly.


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