Question

prove that one of any three consecutive positive integers must be divisible by 3

Answer 1

hey dude....
here is ur answer....!!

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.

thanks.!

Answer 2

Any three consecutive positive integers must be of the form
n, (n + 1)and (n + 2), where n is any natural number, i.e., n = 1,2, 3…
Let,  a = n,b = n+ 1 and c = n + 2
Order triplet is (a, b, c) = (n, n + 1, n + 2), where n = 1,2, 3, …(i)
At n = 1; (a, b, c) = (1,1 + 1,1 + 2) = (1,2, 3)
At n = 2; (a,b,c) = (1,2 + 1,2 + 2) = (2, 3, 4)
At n = 3; (a, b, c) = (3, 3 + 1, 3 + 2) = (3, 4, 5)
At n = 4; (a, b, c) = (4, 4 + 1, 4 + 2) = (4, 5, 6)
At n = 5; (a, b, c) = (5, 5 + 1, 5 + 2) = (5, 6,7)
At n= 8 (a, b, c) = (6, 6+ 1, 6 + 2) = (6,7, 8)
Atn = 7; (a, b, c) = (7,7 + 1,7 + 2) = (7, 8, 9)
At n = 8; (a, b, c) = (8 8 + 1, 8 + 2) = (8, 9,10)
We observe that each triplet consist of one and only one number which is multiple of 3 i.e., divisible by 3.
Hence, one of any three consecutive positive integers must be divisible by 3.

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