Kdtyjb

One needs to choose six real numbers $x_1,x_2,cdots,x_6$ such that the product of any five of them is equal to other number. The number of such choices is

A) $3$

B) $33$

C) $63$

D) $93$

I believe this is not so hard problem but I got no clue to proceed. The work I did till now.

Say the numbers be $a,b,c,d,e, abcde$. Then, $bcdot ccdot dcdot ecdot abcde=a$ hence $bcde= +-1$.

Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?

Well, if there are an even number of “1”s and “-1”s, then the property holds, so that’s $binom{6}{0}+ binom{6}{2}+ binom{6}{4}+ binom{6}{6}=32$. And then there’s the case that they’re all zero. Thus, there are 33 total cases.

Here's an argument which extends to $n$ real numbers quite nicely! We can start by noting that
$$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$
And by commutativity, we get that in fact all the magnitudes are equal.

Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we can conclude that $L=0$ or $1$.

Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s

If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that
$$frac{prod_{i=1}^6 x_i}{1} = 1 text{ and } frac{prod_{i=1}^6 x_i}{-1} = -1$$
Now we can count configurations. There will be
$$sum_{i=0}^3 binom{6}{2i} = 2^5$$
possibilities. And finally, we have $1+32 = 33$.

Just to make the solution more explicit, the product of all six values is the square of any one of them, so they're all $pm k$ for some $k$, and $k^6=k^2implies kin{0,,1}$. We get one $k=0$ solution and, because we require a non-negative product, one $k=1$ solution for each choice of six $pm 1$s with an even number of $-1$s. Famously, the numbers of even- and odd-sized subsets of a finite non-empty set are equal, and so the total number of solutions is $1+frac12cdot2^6$, i.e. B).

HINT:

It is obvious that you need to have the numbers absolute value equal to $1$ (besides one special case but we get to that later). The question is how to set the signs of these $1$s. The product will be negative if you have an odd number of negative signs in it and positive whenever you have an even number of negative signs. Since you are to determine signs for an even number of $1$s if you decide to give an even number of them a negative sign you have one of the following two situations

• the $1$ that is NOT in the product has negative sign. In this case there are a $text{even}-1=text{odd}$ number of negative signs distributed along the other five $1$s which result in $-1$ so the condition is fullfilled




• the $1$ that in NOT in the product has positive sign. In this case there are a $text{even}-0=text{even}$ number of negative signs distributed along the five other $1$s which result in $+1$ so the condition is once again fullfilled.

So your question translates to in how many ways can you distribute an even number of negative signs among the six $1$s. This is solved in the easiest way by considering the number of ways to distribute $0,2,4$ and $6$ negative signs.

Finally the one solution not having to deal with $1$s is the one dealing with all zeros.

Hope this helped