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# A positive integer n is randomly guesses. Let the probability that (xn+1- xn +1) is divisible by (x2 - x +1) is P . Find P

Ribhu Archon
28 Points
one year ago
This question is based on complex numbers so.....

Let’s assume an equation  $x^3=-1$

therefore,

$x^3+1=0$
So all roots of the above cubic equation except 1 are roots of  $x^2-x+1$

Now we know roots of are  $\omega,\omega^2,1$
$\therefore(-x)^3=-1$
So  $(-\omega)^3=-1$    ,    $(-\omega^2)^3=-1$   ,    $(-1)^3=-1$

Clearly roots of the first equation are  $-\omega$   ,  $-\omega^2$   ,$-1$

putting  $\omega$in and simplifying we get, [we can put $-\omega^2$, but i’ll come to that later][step a]

$(\omega)^{n+1}*(-1)^{n+1}+(-1)*(\omega)^{n}*(-1)^{n}+1=0$
$((\omega)^{n})(-\omega^2)(-1)^{n+1}=-1$
[$\omega^2+\omega+1=0\therefore\omega+1=-\omega^2$ ]

$\therefore\omega^{n+2}=(-1)^{n+3}$

Since a power of  $\omega$  is giving a real number, that power must be a multiple of 3.  Hence n+2 is divisible by 3

also the only real result of a power of omega is 1. hence the power of -1 is even. Hence n+3 is divisible by 2

thus n is odd (from n+3 is divisible by 2) and n leaves a remainder 1 when divide by 3 (from n+2 is divisible by 3)

let’s now list out the possibilties,..

${\color{Red} 1}, 2 ,3 ,4 ,5 ,6 ,{\color{Red} 7} ,8 ,9,10,11,12,{\color{Red} 13},14,15,16,17,18,{\color{Red} 19}....$
Notice how every 6 possibilities there is one such number

let the sample space thus be 6*q [q being any integer], there would be q favourable outcomes.
as the sample space is infinite (no restriction given) we can assume that even though the sample space is infinite this ‘infinity’ is divisible by 6 [any other case would be mathematically insignificant]

thus probability is
$P=\lim_{q\to\infty}\frac{q}{6*q}=\frac{1}{6}$

the same will actually happen if in [step a] instead of $-\omega$ we put $-\omega^2$
$(-\omega^2)^{n+1}-(-w^2)^{n}+1=0$

$((\omega^2)^{n})(\omega^2+1)(-1)^{n+1}=-1$
$((\omega^2)^{n})(-\omega)(-1)^{n+1}=-1$ [$\omega^2+\omega+1=0\therefore\omega^2+1=-\omega$]
$((\omega)^{2n+1})(-1)^{n+2}=-1$

$\therefore\omega^{2n+1}=(-1)^{n+3}$

Since a power of  $\omega$  is giving a real number, that power must be a multiple of 3.  Hence 2n+1 is divisible by 3

also the only real result of a power of omega is 1. hence the power of -1 is even. Hence n+3 is divisible by 2

thus n is odd (from n+3 is divisible by 2) and 2n leaves a remainder 2 when divided by 3 (from 2n+1 is divisible by 3)

listing out the possibilities again....

${\color{Blue} 1}, 2 ,3 ,4 ,5 ,6 ,{\color{Blue} 7} ,8 ,9,10,11,12,{\color{Blue} 13},14,15,16,17,18,{\color{Blue} 19}....$
the same result the same equation and thus the same value:

$P=\lim_{q\to\infty}\frac{q}{6*q}=\frac{1}{6}$
hence,

$P=\frac{1}{6}$
Vikas TU
14149 Points
one year ago
Dear student