Not The Brute Force Approach

Algebra Level 5

Let m m be the largest real solution to the equation 3 x 3 + 5 x 5 + 17 x 17 + 19 x 19 = x 2 11 x 4. \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}= x^2-11x-4.

There are positive integers a , b , c a,b,c such that m = a + b + c m = a + \sqrt{b+\sqrt{c}} . Find a + b + c a+b+c .


The answer is 263.

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3 solutions

Aareyan Manzoor
Nov 20, 2015

first off.. i loved this problem, thanks to the creator for giving me such great experience. start like this 3 x 3 + 5 x 5 + 17 x 17 + 19 x 19 + 4 = x 2 11 x 3 x 3 + 1 + 5 x 5 + 1 + 17 x 17 + 1 + 19 x 19 + 1 = x 2 11 x x x 3 + x x 5 + x x 17 + x x 19 = x ( x 11 ) \begin{aligned} \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}+4= x^2-11x\\ \frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1= x^2-11x\\ \frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x(x-11)\end{aligned} we get x=0 as a solution. now divide by x and put x=y+11 1 y + 8 + 1 y 8 + 1 y + 6 + 1 y 6 = y 2 y y 2 64 + 2 y y 2 36 = y \begin{aligned} \frac{1}{y+8}+\frac{1}{y-8}+\frac{1}{y+6}+\frac{1}{y-6}=y\\ \frac{2y}{y^2-64}+\frac{2y}{y^2-36}=y \end{aligned} y=0 ->x=11 is another solution. we divide again to get 2 y 2 64 + 2 y 2 36 = 1 2 ( y 2 36 ) + 2 ( y 2 64 ) = ( y 2 64 ) ( y 2 36 ) 4 y 2 200 = y 4 100 y 2 + 2304 y 4 104 y 2 + 2504 = 0 y 2 = 52 ± 200 y = ± 52 ± 200 x = 11 ± 52 ± 200 \begin{aligned} \frac{2}{y^2-64}+\frac{2}{y^2-36}=1\\ 2(y^2-36)+2(y^2-64)=(y^2-64)(y^2-36)\\ 4y^2-200=y^4-100y^2+2304\\ y^4-104y^2+2504=0\\ y^2=52\pm \sqrt{200}\\ y=\pm\sqrt{52\pm \sqrt{200}}\\ x=11\pm\sqrt{52\pm \sqrt{200}}\end{aligned} we get that x = 0 , 11 , 11 ± 52 ± 200 x=0,11,11\pm\sqrt{52\pm \sqrt{200}} the required x l a r g e s t = 11 + 52 + 200 x_{largest}=11+\sqrt{52+ \sqrt{200}} 11 + 52 + 200 = 263 11+52+200=\boxed{263}

Mathematics is an art of deceiving. I did by twice substitution, first substitute x 3 = a x-3 =a and x 17 = b x-17=b , yielding that a 2 = x 5 a-2 = x-5 and b 2 = x 19 b-2 = x-19 to make it easier. Solving as usual until reaching a b a b ab-a-b form, then substitute into y y and completeing square. Returning to x x and complete square.

I, still, love this deceiving problem

Figel Ilham - 5 years, 6 months ago

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Probably substituting from the start a=x - 11,might be better. (x-3)=a+8, (x-8) =a - 8.,,,,.(x-5)=a+6, (x-17)=a - 6. .........This would simplify Aareyan Manzoor's first stage.

Niranjan Khanderia - 5 years, 5 months ago

Just wondering what is a,b and c as it didn't mention more...

Thaddeus June - 5 years, 6 months ago

Great!! Good solution too....

harish ghunawat - 5 years, 6 months ago

really nice solution but same approach...

Shreyansh Choudhary - 5 years, 6 months ago

For those interested, this was Problem 14 AIME I 2014

Edgar Wang - 5 years, 6 months ago

Same solution

Aakash Khandelwal - 5 years, 6 months ago

Very well done! Did you get any hint from friend?

Lu Chee Ket - 5 years, 6 months ago

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nope, sir. i remember solving this and applied a similar method.

Aareyan Manzoor - 5 years, 6 months ago

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I see. So you mean you learnt from others.

Lu Chee Ket - 5 years, 6 months ago

Enjoyed solving it......

raghav bagri - 5 years, 6 months ago

Same pinch :)

dhiraj agarwalla - 5 years, 5 months ago

All most the same approach. Only used Y = X 2 22 X + 71. S o ( X 3 ) ( X 19 ) = Y 14 , a n d ( X 5 ) ( X 17 ) = Y + 14. E q u a t i o n t u r n s t o b e Y 2 4 Y 1 4 2 = 0 Solving we get Y. Substituting for Y, and solving for X we get the answer. Y=X^2 - 22X +71. ~~~~~~ ~ So ~ (X - 3)(X - 19)=Y - 14, ~ and ~(X - 5)(X - 17)=Y + 14.\\ Equation ~ turns ~ to ~ be ~Y^2 - 4Y - 14^2=0\\ \text{Solving we get Y. Substituting for Y, and solving for X we get the answer.}

Niranjan Khanderia - 5 years, 5 months ago

You wrote it perfectly. Thumbs up!

Ais V - 5 years, 5 months ago

Did the same way.

Anupam Nayak - 5 years, 5 months ago
Lu Chee Ket
Nov 23, 2015

General method should expand b l i n d l y blindly but not too difficult with Excel. { A B r u t e F o r c e A p p r o a c h ! A Brute Force Approach! }

x 5 55 x 4 + 1106 x 3 9878 x 2 + 37957 x 50171 = 0 x^5 - 55 x^4 + 1106 x^3 - 9878 x^2 + 37957 x - 50171 = 0 {An x = 0 was eliminated.}

Figures of Excel correlated with other 18 S.F. solver:

α \alpha = 11.00000000000000

β \beta = 19.13278154285057

γ \gamma = 17.15287448078287

δ \delta = 4.847125519217132

ϵ \epsilon = 2.867218457149427

All checked to be correct to very close's proximity.

Numerical method can achieve a merit easily for 19.13278154285057 \rightarrow 11 11 + 52 + 200 \sqrt{52 + \sqrt{200}} .

11 + 52 + 200 = 263

Answer: 263 \boxed{263}

Lu, to save your trouble. Just use Wolfram Alpha ( see here )but computer solution is unneeded here. Click More roots under Solutions.

Chew-Seong Cheong - 5 years, 6 months ago
Andreas Wendler
Dec 19, 2015

Very easy and fast one gets the solution via Xmaxima:

solve(x^2-11*x-4-3/(x-3)-5/(x-5)-17/(x-17)-19/(x-19),x);

                          7/2                     7/2
            sqrt(208 - 5 2   )      sqrt(208 - 5 2   )
  [x = 11 - ------------------, x = ------------------ + 11, 
                    2                       2

                    7/2                     7/2
            sqrt(5 2    + 208)      sqrt(5 2    + 208)
   x = 11 - ------------------, x = ------------------ + 11, x = 11, x = 0]
                    2                       2

As third last value of the solutions' list you find the largest real solution (x=19.13278...) and after some equivalent transformation you have x = 11 + 52 + 200 x = 11 + \sqrt{ 52 + \sqrt{ 200 }} .

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