Question

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

Answer 1

Let the the three consecutive numbers be n, n+1 ,n+2 .

Checking the argument :- Only one of n, n+1,n+2 would be divisible by 3 when n is +ve integer. 

Case 1 :- 

n is divisible by 3 .

then n+2/3 = n/3 + 2/3 this shows a remainder 2 .

then n+1/3 =1+ n/3 + 1/3 this shows a remainder 1.

Case 2 :- 

n+2 is divisible by 3 .

n/3 = n+2-2/3 = n+2/3 - 2/3 this leaves a remainder (3-2) = 1 .

n+1/3 =n+2-1/3 = n+2/3 + 2/3 leaves a remainder 2 .

Case 3 :- 

n+1 is divisible by 3 .

n/3 = n+1-1 /3 = (n+
1)/3 - 1/3 this leaves a remainder (3-1) = 2.

n+2/3 =n+1 + 1 /3 = n+4/3 + 1 /3 leaves a remainder (3-2)= 1 .

Hence proved that; one and only one out of n, n+1, n+2 is divisible by 3, where n is any positive integer..

From this, We can conclude that alteast 1 of three consecutive numbers is divisible by 3

Answer 2

Step-by-step explanation:

Let 3 consecutive positive integers be n, n + 1 and n + 2 .

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

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

If n = 3p = 3(p) , then n is divisible by 3

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

Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3

Hence it is solved.

THANKS

#BeBrainly.