Question

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

Answer 1

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

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Answer 2

always, the three consecutive terms will be x-1,x,x+1
if two of these tems are declared to be undivisible by 3, naturally the leftover term will be a multiple of 3
1.when x+1,x are not multiples of 3, x-1 will be a multiple.eg:3,4,5
2.when x-1,x are not multiples of 3, x+1 will be a multiple.eg:4,5,6
3.when x-1,x+1are not multiples of 3, x will be a multiple.eg:2,3,4