Submitted by Francis Gerard Magtibay

Algebra Level 4

Suppose $x+ \frac {1}{x}= \frac{4 \sqrt {3}}{3}$ . Then $x^{10}+ \frac {1}{x^{10}}=a+ \frac {b}{c}$ , where $a$ , $b$ and $c$ are positive integers, $\frac{b}{c}$ is less than $1$ , and $b$ and $c$ are coprime. Find $a+b+c$ .

This problem is from the UP Math Wizard 2010.

The answer is 487.

9 solutions

Kaustav Mukherjee
Dec 17, 2013

First, make the given equation $x+\frac{1}{x}$ quadratic by multiplying /(x/) throughout the equation.

we get, $3x^{2} -4\sqrt{3}x+3=0$

Now, let us find the roots of the equation

$x=4\sqrt{3} +\sqrt{(4\sqrt{3})^{2} - 4.3.3)}$ and $x=4\sqrt{3} -\sqrt{(4\sqrt{3})^{2} - 4.3.3)}$

we get the two solution, $x=\sqrt{3} , \frac{1}{\sqrt{3}}$

Now put the any value of x to the equation $x^{10} + \frac{1}{x^{10}}$

we get answer as $243 +\frac{1}{243}$

Hence, we find a,b,c

Therefore, $a+b+c=\boxed{487}$

You can also get a different answer at the question mentions that b and c are coprimes. Answer is 243 + 1/243. If you manipulate a little, you will get 242+ 244/243, in which case the answer will be 729(243 and 244 still being coprimes). Another answer below 999 would be 971(241 + 487/243). What say guys...

Prerak Shah - 7 years, 4 months ago

Yeah, I didn't notice that. I'll rephrase the question to avoid this loophole. Thanks. :)

Francis Gerard Magtibay - 6 years, 3 months ago

Wow..I never thought this is so easy.

Jayakumar Krishnan - 6 years, 9 months ago

its very fine

seenu mandadi - 7 years, 5 months ago

gud1

Math Lovers - 7 years, 5 months ago

升泽林
Dec 17, 2013

First find the x

$x + \frac{1}{x}=\frac{4\sqrt{3}}{3}$ times x ，get $x^{2}-\frac{4\sqrt{3}}{3}x +1=0 =(x-\sqrt{3})(x-\frac{\sqrt{3}}{3})=0$ $x=\sqrt{3} or \frac{\sqrt{3}}{3}$

Then substitute to $x^{10}+\frac{1}{x^{10}}$ get the same answer $3^{5}+\frac{1}{3^{5}}$

$a+b+c = 3^{5}+1+3^{5} =487$

Fahad Hridoy
Dec 17, 2013

x+1/x=4/√3 =√3+1/√3 so,we get the value of x,that is √3 now,(√3)^10+(1/√3)^10 =3^5+1/3^5 =243+1/243 so,a=243 b=1 c=243 so,a+b+c=243+1+243=487

Shikhar Jaiswal
Feb 16, 2014

by observation:x=3^(0.5)

Alvarana Elvatra
Dec 19, 2013

x + 1/x = 4(3^0.5)

(x + 1/x)^2 = (4(3^0.5))^2

x^2 + 2 + 1/x^2 = 16/3

x^2 + 1/x^2 = 16/3 -2

                  = 16/3 - 6/3

                  = 10/3

                  = 3 + 1/3

x^2 = 3 1/x^2 = 1/3

x = 3^0.5

x^10 + 1/x^10 = a + b/c

(3^0.5)^10 + 1/(3^0.5)^10 = 243 + 1/243

a = 243 b = 1 c = 243

a+b+c = 243 +1 + 243 = 487

Vipul Panwar
Dec 18, 2013

(x+1/x)^2=x^2+(1/x^2)+2 so x+1/x=4/root(3) so x^2+(1/x^2)=10/3 x^4+(1/x^4)=82/9 x^8+(1/x^8)=6562/81 (x^3+(1/x^3))^2=784/27 x^10+(1/x^10)= (x^8+(1/x^8))(x^2+(1/x^2))-(x^3+(1/x^3))^2+2 so x^10+(1/x^10)=243+1/243 therefore a=243 b=1 c=243 a+b+c= 487

Budi Utomo
Dec 17, 2013

x + 1/x = (4 . 3^1/2)/3 = (3.3^1/2)/3 + (3^1/2)/3 = 3^1/2 + 1/3^1/2. So, x = 3^1/2 . x^10 + 1/x^10 = (3^1/2)^10 + 1/(3^1/2)^10 = 3^(1/2 . 10) + 1/(3^(1/2.10) = 3^5 + 1/3^5 = 243 + 1/ 243 So a + b/c = 243 + 1/243 ----------------> a + b+ c = 243 + 1 + 243 = 487.

Suvam Das
Dec 17, 2013

calculate x-1/x to obtain value of x (p.s. the positive square root).

Jade Nicholas Fabella
Dec 17, 2013

first step is to combine x + 1/x into (x^2 +1 )/x .. after doing that, square the equation and you will get (x^4+2x^2+1)/x^2=16/3.. if you simplify it, you will get 3x^4 -10x^2 +3=0 .. factor out and you will get sqrt(3) and sqrt(1/3) as two positive roots.. substitute the following values to x^10 + 1/x^10 and you will get 59050/243 or 243 + 1/243 ...

thus; 243 + 243 + 1 = 487 :))

How did you simplify (x^4+2x^2+1)/x^2=16/3 to 3x^4 -10x^2 +3=0? and how to get the roots faster from 3x^4 -10x^2 +3=0? i mean i can't imagine the roots are from factorial mode simply.

Hafizh Ahsan Permana - 7 years, 5 months ago

Submitted by Francis Gerard Magtibay

The answer is 487.

9 solutions

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