Question

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

Answer 1

Let the positive integers be a, a+1, a+2
We know that every positive integer "a" is of the form 3m, 3m+1, 3m+2
We will have three cases:
CASE 1 : a=3m
=>a is divisible by 3 but a+1 and a+2 are not.

CASE 2 : a=3m+1
=>a+2 is divisible by 3 but a and a+1 are not.

CASE 3 : a=3m+2
=>a+1 is divisible by 3 but a and a+2 are not.

By all above cases we can clearly see that out of every three consecutive positive integers one integer is divisible by 3.

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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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