Math, asked by khushi02022010, 8 months ago

Prove that own and only one out of any three consecutive positive integers is divisible by 3.​

Answers

Answered by Anonymous
5

Let the three consecutive positive integers be n,n+1,n+2

We know that n is of the form 3q,3q+1 or, 3q+2

Hence we can consider the following three cases:

Case I When n=3q n is divisible by 3 but it is not possible for n+1 and n+2

Case II When n=3q+1 n+2=3q+1+2=3 is divisible by 3 but it is not possible for n and n+1

Case III When n=3q+2 n+1=3q+1+2=3(q+1) is divisible by 3 but it is not possible for n and n+2

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

Answered by Anonymous
10

SOLUTION :)

Let the three consecutive positive integers be n,n+1,n+2

We know that n is of the form 3q,3q+1 or, 3q+2

Hence we can consider the following three cases:

Case I When n=3q n is divisible by 3 but it is not possible for n+1 and n+2

Case II When n=3q+1 n+2=3q+1+2=3 is divisible by 3 but it is not possible for n and n+1

Case III When n=3q+2 n+1=3q+1+2=3(q+1) is divisible by 3 but it is not possible for n and n+2

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

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