Solve for n to prove that the given equation has a solution.

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In summary, to solve for n in a given equation, you need to isolate it using algebraic operations and then plug in different values to see if they satisfy the equation. Any value for n can be used to prove the equation has a solution, but some values may be more helpful than others. If no value of n satisfies the equation, then the equation has no solution. An equation can have one, two, or infinite solutions depending on its complexity and variables. Proving that an equation has a solution may not always be necessary, but it can aid in understanding the relationship between variables and solving related problems.
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Chris L T521
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I would like to thank those of you that participated in the first POTW on MHB. Now it's time for round two! (Bigsmile)This week's problem was proposed by caffeinemachine.Problem: Let each of the numbers $a_1, a_2, \ldots , a_n$ equal $1$ or $-1$. If we're given that $a_1 a_2 a_3 a_4 + a_2 a_3 a_4 a_5 + \ldots + a_n a_1 a_2 a_3 =0$, prove that $4|n$.There were no hints provided for this problem.Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-(POTW)-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!EDIT: Since there was some confusion as to what this was asking, I included a clarification in the spoiler.

Here are some more terms in the sum: $a_1a_2a_3a_4 + a_2a_3a_4a_5 + a_3a_4a_5a_6+a_4a_5a_6a_7+\cdots+a_{n-2}a_{n-1}a_na_1 + a_{n-1}a_na_1a_2+a_na_1a_2a_3$.

You may now be asking what happens in the case of $n=4$. Well, if you follow this pattern, you are cycling through the indices 4 at a time, and once you get to $n$, you "start over again." So, if $n=4$, the indices of the terms in the sum will be a cyclic permutation of $(1234)$ [starting with $(1234)$ and ending with $(4123)$].
 
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Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution:

[sp]Claim: $n$ is even.
Proof: Since each of $a_i's$ is either $1$ or $-1$ each of the summands in the LHS of the given equation is also either $1$ or $-1$. Since the RHS is $ 0$ the number of $1's$ appearing in the LHS is equal to the number of the $-1's$ appearing in the LHS. Thus $2|n$.

Claim: $4|n$.
Proof: From the above claim we can write $n=2k$.
Consider $(a_1a_2a_3a_4)(a_2a_3a_4a_5) \ldots (a_na_1a_2a_3)=(a_1a_2 \ldots a_n)^4$ .
In the above RHS is definitely $1$ while LHS is $(-1)^k (1)^k$.
Thus $(-1)^k(1)^k=1 \Rightarrow (-1)^k=1 \Rightarrow 2|k.$ Hence $4|n$.[/sp]
 

Related to Solve for n to prove that the given equation has a solution.

1. How do I solve for n to prove that the given equation has a solution?

To solve for n, you need to isolate it on one side of the equation by using algebraic operations such as addition, subtraction, multiplication, and division. Once you have n on its own, you can plug in different values to see if they satisfy the equation.

2. Can I use any value for n to prove that the given equation has a solution?

Yes, you can use any value for n as long as it satisfies the equation. However, some values may be more useful in proving that the equation has a solution than others.

3. What if I can't find a value for n that satisfies the equation?

If you have exhausted all possible values for n and none of them satisfy the equation, then the equation does not have a solution. This means that there is no value of n that makes the equation true.

4. How many solutions can an equation have?

An equation can have one, two, or infinite solutions. It depends on the complexity of the equation and the number of variables involved.

5. Is it necessary to prove that an equation has a solution?

No, it is not always necessary to prove that an equation has a solution. Some equations are already known to have solutions, while others may not have a solution at all. However, proving that an equation has a solution can help in understanding the relationship between the variables and can also be useful in solving other related problems.

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