Question about arithmetic

[inverse]

Hello,

The formulation of the question says:

Show, using congruences and disjunction of cases, which,for all natural n є Ninteger, the integern (n+1) (2n+1) is a multiple of 6.

Simultaneously, themultiple of 3 is a multiple of 2 (number), then it is divisible by 6

3 cases are possible by applying the congruence module 3:

If $n=3k \Longrightarrow n 3k\left ( n+1 \right )$$\left ( 2n+1 \right )=3k'$

If $n=3k+1 \Longrightarrow 2n+1=6k+3$ $3k\left ( n+1 \right )\left ( 2k+1 \right )=3k''$

If [itex]n=3k+2 \Longrightarrow n+1=3k+3[/tex] and [tex]3n\left ( k+1 \right )\left ( 2n+1 \right )=3k'''[/itex]

Thank you very much for your attention

 

[DonAntonio]

Hello,

The formulation of the question says:

Show, using congruences and disjunction of cases, which,for all natural n є Ninteger, the integern (n+1) (2n+1) is a multiple of 6.

Simultaneously, themultiple of 3 is a multiple of 2 (number), then it is divisible by 6

3 cases are possible by applying the congruence module 3:

If $n=3k \Longrightarrow n 3k\left ( n+1 \right )$$\left ( 2n+1 \right )=3k'$

If $n=3k+1 \Longrightarrow 2n+1=6k+3$ $3k\left ( n+1 \right )\left ( 2k+1 \right )=3k''$

If [itex]n=3k+2 \Longrightarrow n+1=3k+3[/tex] and [tex]3n\left ( k+1 \right )\left ( 2n+1 \right )=3k'''[/itex]

Thank you very much for your attention

You should be more careful before you send over a post, and that's what the "preview post" button exists for.

You want to know for what positive integers n, the number $\,(n+1)(2n+1)$ is a multiple of 6.

Your division in cases is fine, but for example: $$n=3k\Longrightarrow (n+1)(2n+1)=(3k+1)(6k+1)$$ which is never a multiple of 6 (I just can't understand what you did in this case)

Simmilarly for the next case, $$n=3k+1\Longrightarrow (n+1)(2n+1)=(3k+2)(6k+3)$$ which is multiple of 6 iff $\,k\,$ is even (why?).

Now try to do correctly the last case by yourself.

DonAntonio

 

[inverse]

I cannot understand what are you asking me, can you ask the question again?
thanks

 

[ramsey2879]

I cannot understand what are you asking me, can you ask the question again?
thanks

Don asked you to check your first post for errors and typos using the preview button. Looks like you should have corrected the typos as he only saw the last two factors since you left out a space between "integers" and "n". By the way you did not show that the product is divisible by 2, so your work is incomplete.

 

[blue_raver22]

and your question. I am a scientist and I specialize in a different field than mathematics, so I may not be able to provide a comprehensive response to your question. However, I can provide some insights based on my knowledge and understanding of arithmetic.

The question you have presented involves the concept of congruences and disjunction of cases. Congruence is a mathematical concept that is used to compare two numbers and determine if they have the same remainder when divided by a given number. In this case, you are trying to show that for all natural numbers n, the expression (n+1)(2n+1) is a multiple of 6. This means that when divided by 6, the expression will have a remainder of 0.

To prove this, you have broken down the problem into three cases based on the value of n modulo 3. In the first case, when n is a multiple of 3, the expression can be rewritten as 3k(n+1)(2n+1), where k is a natural number. This clearly shows that the expression is a multiple of 3. Similarly, in the second and third cases, when n is not a multiple of 3, the expression can be rewritten as 3k'(n+1)(2n+1) and 3k''(n+1)(2n+1), respectively. In both cases, the expression is still a multiple of 3.

Furthermore, since the question also mentions that the multiple of 3 is a multiple of 2, we can conclude that the expression (n+1)(2n+1) is a multiple of 6. This is because when a number is a multiple of both 2 and 3, it is also a multiple of their product, which is 6.

In conclusion, through the use of congruences and disjunction of cases, we have shown that for all natural numbers n, the expression (n+1)(2n+1) is a multiple of 6. I hope this helps to clarify your question. If you have any further inquiries, please feel free to ask. Thank you again for your interest in arithmetic and congruences.