Question

prove that one out of every three consecutive positive integer is divisible by 3

Answer 1

The proof is simple. Let the three consecutive positive integers be x, x+1, and x+2 where x is a positive integer. 
Now, when a number is divided by 3, it leaves three possible remainders: 0, 1, and 2. 
Remembering that Dividend= Divisor x Quotient + Remainder,
Thus, the number x can be either of the form 3n (if it leaves no remainder), 3n+1 (if it leaves remainder 1), or 3n+2 (if it leaves remainder 2), where n is a positive integer.
Case I: 
Now, if x=3n, one of the three original numbers is divisible by three.


Case II:

If x=3n+1, then the third number in series, x+2 would be 3n+3, and will be divisible by three.

Case III:

If x=3n+2, the second number x+1 would be 3n+3, and will be divisible by three. 

Hence, in either of the cases, whatever x might be, one of the three consecutive numbers would always be 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.

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