Math, asked by boulzhi, 1 year ago

prove that one an only one out of n,n+2 and n+4 i divisible by 3 where n is any positive integer

Answers

Answered by ADRIJAjayant

Let n be equal to 2.

1)Verifying n/3,

Answer:2/3=0.66

Therefore, it is not divisible by 3.

2)Verifying n+2/3

Answer:2+2/3

=4/3

=1.33

Therefore, it is not divisible by 3.

3)Verifying n+4/3

Answer:2+4/3

=6/3

Therefore, this is the ONLY equation where it is fully divisible by 3.

HannesSidorov: Although your example did work here, this is not exactly a correct proof. You have to prove it to be true generally, not for just one specific number, in this case 2.

ADRIJAjayant: i know but while proving, you generally assume the value and if it implies to one integer, it also implies to the others

HannesSidorov: Well, that doesn't necessarily have to be true. A single example is still not enough to prove the thing, however logical that might seem.

Answered by HannesSidorov

Let's look at three possible cases:

1) n=3k,

2) n=3k+1

3) n=3k+2,

where k is some non-negative integer.

1) If n=3k, then we get three numbers 3k, 3k+2 and 3k+4. 3k is divisible by 3, 3k+2 is not divisible by 3 and 3k+4=3(k+1)+1 is also not divisible by 3.

2) If n=3k+1, then we get three numbers 3k+1, 3k+3 and 3k+5. 3k+1 and 3k+5 are not divisible by 3, while 3k+3 is divisible by 3.

3) If n=3k+2, then we get three numbers 3k+2, 3k+4 and 3k+6. Numbers 3k+2 and 3k+4 are not divisible by 3, while 3k+6 is divisible.

Since in all three possible cases there was one number out of three that was divisible by three, the claim is true and proven.

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