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

1. $3$
2. $33$
3. $63$
4. $93$
edited | 580 views
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option A
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how??
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(1,1,1,1,1,1) , (0,0,0,0,0,0), (-1,-1,-1,-1,-1,-1)
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+1
Yes, answer should be $33$ only
+1

My approach is similar to  Kushagra 's approach

So, as per the question, we're choosing six real numbers

namely, $X_1,X_2,X_3,X_4,X_5,X_6$

Now, I'll write $6$ equations that should be satisfied the criteria:

The product of any five of them is equal to other number.

$$X_1.X_2.X_3.X_4.X_5=X_6 ---- A) \\ X_1.X_2.X_3.X_4.X_6=X_5 ---- B) \\ X_1.X_2.X_3.X_5.X_6=X_4 ---- C) \\ X_1.X_2.X_4.X_5.X_6=X_3 ---- D) \\ X_1.X_3.X_4.X_5.X_6=X_2 ---- E) \\ X_2.X_3.X_4.X_5.X_6=X_1 ---- F)$$

Now, take A) & B) & divide them

$\dfrac{A)}{B)}$     $:$   $\dfrac{X_1.X_2.X_3.X_4.X_5}{X_1.X_2.X_3.X_4.X_6} = \dfrac{X_6}{X_5}$

Or,   $\dfrac{A)}{B)}$  $:$   $\dfrac{X_5}{X_6} = \dfrac{X_6}{X_5}$

Or, $(X_5)^2 = (X_6)^2$

take A) & C) & divide them,

$\dfrac{A)}{C)}$     $:$   $\dfrac{X_1.X_2.X_3.X_4.X_5}{X_1.X_2.X_3.X_5.X_6} = \dfrac{X_6}{X_4}$

Or,   $\dfrac{A)}{B)}$  $:$   $\dfrac{X_4}{X_6} = \dfrac{X_6}{X_4}$

Or, $(X_4)^2 = (X_6)^2$

∴ If we take any pair from A), B), C), D), E), F) and divide one by other, we'll get -

$$(X_1)^2 = (X_2)^2 = (X_3)^2 = (X_4)^2 = (X_5)^2 = (X_6)^2$$

∴ All the numbers have the same magnitude, only difference will be their sign.

Let's assume the magnitude of each of the real numbers is $\pm X$.

So, the arrangement of plus or minus signs is going to determine the arrangements possible there.

And we have to find how we'll get

$$X.X.X.X.X = X \\ X^5 = X$$

Only when $x =1$ or $X=0$, then $X^5 = X$

But, there could be an another way to get $X^5 = X$

When there is even number of $-1's$ (i.e. $2$ or $4$ only), the product of $5 x$ will be $1$ which is equal to the selected $x$ (i.e. $1$)

[number of $-1's$ can't be $6$ because, product of $5$ real no.s should be equal to the $6^{th}$ no.(other number)]

$$-1.-1.1.1.1 = 1 \\ \text{ Or} \\ -1.-1.-1.-1.1 =1$$

Now, if we switch to the binary system where

$$\text{0 represents +} \\ \text{1 represents -}$$

Then we have $2^6 = 64$ combinations of plus & minus

but, we only want those combination where we have only even no. of $-1's$

We can apply the parity concept.

Now, we have only $32$ combinations of bits with even parity.

So, we have $32$ arrangements for the $x's$ with their signs which satisfy the criteria or the required conditions.

plus we have one more solution where all $x's$ are $0$.

∴ $\color{green}{\text{Total arrangement or choices will be} = 32+1}$ = $\color{gold}{33}$

& this can think of-

1) Where all $x's$ are $0$, product of $5$ $x's$ will be $0$, which is equal to another $x$ (i.e. $0$)

2) Where all $x's$ are $+1$, product of $5$ $x's$ will be $+1$, which is equal to another $x$ (i.e. $+1$)

3) Where all $x's$ are $-1$

$$\text{think as, } -1.-1.-1.-1.-1 = -1$$

4) Where $4$ $x's$ are $+1$ & $2$ $x's$ will be $-1$

$$\text{think as, } +1.+1.+1.+1.-1 = -1$$

5) Where $4$ $x's$ are $-1$ & $2$ $x's$ will be $+1$

$$\text{think as, } -1.-1.-1.-1.+1 = +1$$

1) can be done in = $1$ ways  $\left [ \dfrac{6!}{6!} = 1 \right]$

2) can be done in = $1$ ways  $\left [ \dfrac{6!}{6!} = 1 \right]$

3) can be done in = $\dfrac{6!}{6!} = 1 ways$

4) can be done in = $\dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2} = 15\text{ways}$

5) can be done in $\dfrac{6!}{4! \times 2!} = \dfrac{6 \times 5}{2} = 15\text{ways}$

∴ $\color{green}{\text{Total choices }}$= $1) + 2) + 3) + 4) + 5)$

$\qquad \qquad = 1+1+1+15+15$

$\qquad \qquad = \color{gold}{33 \text{ choices }}$

edited

Here we have to choose 6 real numbers in such a way that the product of any 5 numbers is equal to the 6th number.

So, we will have x1x2x3x4x5 = x6  .............................(1)

and x1x2x3x4x6 = x5 ............................. (2)

From equation 1 we get x1x2x3x4 = x6/x5  ………………….(3)

And from equation 2 we get x1x2x3x4 = x5/x6 ……………....(4)

From equation 3 and 4 we get (x6/x5)2 = 1

=>  x62 = x52

=> |x6| = |x5|

In this way we can prove that the absolute values of all the xi’s are equal

Let |xi| = a

Equation 1 can be written as |x1x2x3x4x5| = |x6|

=> |x1||x2||x3||x4||x5| = |x6|

=> a5 = a

=> a = 1 ( a can be equal to zero also I will consider it later on)

So, xi can be 1 or -1

Now, assume we have chosen (x1,x2,x3,x4,x5,x6) in such a way that

x1 x2 x3 x4 x5 = x6 ……………….(5)

=> (x1 x2 x3 x4) (1/x6) = (1/x5) …………(6)

Since xi can be 1 or -1 so  (1/xi) = xi

Hence x6 = 1/x6 and x5 = 1/x5

So, equation (6) becomes

(x1 x2 x3 x4 x6) = x5

So, if we can choose a 6 tuple (x1,x2,x3,x4,x5,x6) in such a way that equation (5) is true and all the xi’s are either 1 or -1 then the product of any 5 of them becomes equal to the other number.

Now, here x1,x2,x3,x4,x5 have 2 options each they can be either 1 or -1 and x6 is dependent on other 5 numbers.

So, the number of 6 tuples (x1,x2,x3,x4,x5,x6) satisfying equation (5) is 25 = 32

Here we have left the 6-tuple (0,0,0,0,0,0)

So, the number of ways to form 6-tuples such that the product of any 5 of them is equal to the other one is 32 +1 = 33

Option B is the correct answer.

+4

There are 2 cases(as proved above) -
1) All $x_i's = 0$
2) $x_i = 1 \ or \ -1$

For case 1-
Only 1 combination is possible - $(0,0,0,0,0,0).$

For case 2 -
Take 2 examples -
i) $\{x_1,x_2,x_3,x_4,x_5,x_6\} = \{1,-1,-1,1,1,1\}$
Note - The cardinality of -1 is even here. I can write any $x_i$  as a product of other $5$ real numbers.

ii) $\{x_1,x_2,x_3,x_4,x_5,x_6\} = \{1,-1,-1,-1,1,1\}$
Note - The cardinality of $-1$ is odd here. I can't write any $x_i$  as a product of other $5$ real numbers.
Suppose I take $x_2$.
$x_2 = -1$ but the product of other 5 real numbers is 1.
If I take $x_1$, which is 1, but the product of other 5 real numbers is -1.

It's clear from the above two examples that the cardinality of -1's should be even.

Now imagine these six real numbers as a string consists of an even number of 1's and - 1'

Total number of choices for these six real numbers =
Number of choices for Case 1 + Number of choices for Case 2

= 1 + (All 1's) + (two -1's and four 1's)+ (four -1's and two 1's) +(All -1's)

$= 1 + 1 + \frac{6!}{2! \ 4!} + \frac{6!}{2! \ 4!} + 1$

$= 1+1+15+15+1$

$= \color{Red}{33}.$