Redirected
reopened by
2,616 views
1 votes
1 votes

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. $93$
reopened by

3 Answers

Best answer
8 votes
8 votes

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 by
3 votes
3 votes

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.

1 votes
1 votes
x1, x2, x3, x4, x5, x6 can take 0, 1 or -1.

Now one case is where all x takes 0 values => 1 combination

Now, based on -1, we can assign -1 to 0 x’s(all x’s are one), or 2 x’s or 4 x’s or 6 x’s

when 0 x’s are -1 => 1 combination

when 2 x’s are -1 => 6C2 = 15 combinations

when 4 x’s are -1 => 6C4 = 15 combinations

when 6 x’s are -1 => 6C6 = 1 combination

So a total of 1 + 1 + 15 + 15 + 1 = 33 combinations possible

Related questions

1 votes
1 votes
1 answer
1
akash.dinkar12 asked May 11, 2019
947 views
Number of real solutions of the equation $x^7 + 2x^5 + 3x^3 + 4x = 2018$ is$1$$3$$5$$7$
2 votes
2 votes
1 answer
2
akash.dinkar12 asked May 11, 2019
717 views
The sum of the infinite series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots $ is$2$$3$$4$$6$
0 votes
0 votes
1 answer
3
akash.dinkar12 asked May 11, 2019
675 views
The $x$-axis divides the circle $x^2 + y^2 − 6x − 4y + 5 = 0$ into two parts. The area of the smaller part is$2\pi-1$$2(\pi-1)$$2\pi-3$$2(\pi-2)$
0 votes
0 votes
1 answer
4
akash.dinkar12 asked May 11, 2019
728 views
The angle between the tangents drawn from the point $(1, 4)$ to the parabola $y^2 = 4x$ is$\pi /2$$\pi /3$$\pi /4$$\pi /6$