Determining which statement is true

[ver_mathstats]

We are given several statements and we must determine which one is true, n is a fixed positive integer, it asks which statement is true regardless of which positive integer n you choose?

The first one states that if n is divisible by 6, then n is divisible by 3.
I chose 12 for n and it was true.
I chose 18 for n and it was true as well.
Does this mean that the statement is true? Because I believe it is true.

Another one states if n is divisible by 3, then n is divisible by 6.
This one is still true because if you substitute 12 and 18 it still holds true.

Another one also states that if n² is divisible by 6, then n is divisible by 6.
I chose 6 for n and it held true, ((6)²)/6 and 6/6 worked out meaning it must be true.

Is this the correct way to solve this question or am I doing it incorrectly?

Thank you.

 

[stockzahn]

We are given several statements and we must determine which one is true, n is a fixed positive integer, it asks which statement is true regardless of which positive integer n you choose?

Is this the correct way to solve this question or am I doing it incorrectly?

Good morning,

first of all, if this is some kind of homework please use the homework forum with its template. Secondly, it doesn't suffice to prove a mathematical statement with two examples, so I think you should find a proof, which applies for the infinity of numbers (or one converse example for disprooving).

The first one states that if n is divisible by 6, then n is divisible by 3.
I chose 12 for n and it was true.
I chose 18 for n and it was true as well.
Does this mean that the statement is true? Because I believe it is true.

Assumption: $n\;mod\;6 = 0 \rightarrow n\;mod\;3 = 0$
$6 = 2\cdot 3$
$\frac{n}{6} = \frac{n}{2\cdot 3} = \frac{1}{2}\frac{n}{3} \rightarrow \frac{n}{3}\;mod\;2 = 0$
The modulo of a number only can be zero, if it is integral, therefore $\frac{n}{3}$ must be an integral and $n\;mod\;3 = 0$

Another one states if n is divisible by 3, then n is divisible by 6.
This one is still true because if you substitute 12 and 18 it still holds true.

What about the $n=9$?

Another one also states that if n² is divisible by 6, then n is divisible by 6.
I chose 6 for n and it held true, ((6)²)/6 and 6/6 worked out meaning it must be true.

Try by yourself. Find a chain of assumptions and proofs or one disprooving example to verify or falsify the assumption.

 

[ver_mathstats]

Good morning,

first of all, if this is some kind of homework please use the homework forum with its template. Secondly, it doesn't suffice to prove a mathematical statement with two examples, so I think you should find a proof, which applies for the infinity of numbers (or one converse example for disprooving).
Assumption: $n\;mod\;6 = 0 \rightarrow n\;mod\;3 = 0$
$6 = 2\cdot 3$
$\frac{n}{6} = \frac{n}{2\cdot 3} = \frac{1}{2}\frac{n}{3} \rightarrow \frac{n}{3}\;mod\;2 = 0$
The modulo of a number only can be zero, if it is integral, therefore $\frac{n}{3}$ must be an integral and $n\;mod\;3 = 0$
What about the $n=9$?
Try by yourself. Find a chain of assumptions and proofs or one disprooving example to verify or falsify the assumption.

I was unaware of the homework template, I shall start using that now, my apologies. Thank you for the help. I am going to retry the question right now.

 

[fresh_42]

I was unaware of the homework template, I shall start using that now, my apologies. Thank you for the help. I am going to retry the question right now.

The homework template is automatically inserted when you post in the homework forum:
https://www.physicsforums.com/forums/calculus-and-beyond-homework.156/
where your questions actually should have been posted.

 

[ver_mathstats]

The homework template is automatically inserted when you post in the homework forum:
https://www.physicsforums.com/forums/calculus-and-beyond-homework.156/
where your questions actually should have been posted.

Okay thank you, I found it now, my apologies.