Question

prove that one and only one out of three consecutive natural number is divisible by 3​

Answer 1

Step-by-step explanation:

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.

Answer 2

Answer:

since n,n+1,n+2 are three consecutive integers then there must be one number divisible by 3 at least .