Question

Prove that one of any three consecutive
positive integers must be divisible by 3.​

Answer 1

$\huge \sf \blue{Answer}$

${ \textsf {\textbf{if \: n = 3p + 1 \: then \: n + 2 = 3p + 2 + 1 = 3p + 3 = 3}}}$

${ \textsf {\textbf{so \: we \: can \: say \: that \: one \: of \: the \: numbers}}}$

${ \textsf { \textbf{among \: n \: n + 1 \: n + 2 \: is \: always \: divisible \: by \: 3}}}$

Answer 2

$\large\underline\mathfrak\red{Answer \: :- }$

Let n , n+1 , n+2 be three consecutive positive integer.

We know that,

n is the form of 3q , 3q+1 , 3q+2 ( As per Euclid Division Lemma ) ,

So, we have the following

Case I

When n = 3q

In this case, n is divisible by 3 but n+1 and n+2 are not divisible by 3.

Case II

When n = 3q+1

In this case, n + 2 = 3q + 1 + 2 = 3( q + 1 ) is divisible by 3 but n and n+1 are not divisible by 3.

Case III

When n = 3q+2

In this case, n+1 = 3q + 1 + 2 = 3(q+1) is divisible by 3 but n and n+2 are not divisible by 3.

Hence, one of n , n+1 , n+2 is divisible by 3.

$\large\bf\underline\pink{Hence \: Proved \: ! }$

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