Math, asked by hefnarose, 1 year ago

Prove that one of every three consecutive positive integer is divisible by 3

Answers

Answered by TheStudiousBoy27
21

any three positive integer

=x+2x+3x

=6x

hence 6x is divisible by three

therefore any one of every three consecutive positive integer is divisible by 3


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Answered by 34545
25

Answer:

Step-by-step explanation:

Let three consecutive positive integers be n, =n + 1 and n + 2

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. 

If n = 3p, then n is divisible by 3. 

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 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

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. 

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