100,000 points problem

Algebra Level 5

Let m m be the the largest real solution to the equation

3 x 3 + 5 x 5 + 17 x 17 + 19 x 19 = x 2 11 x 4 \dfrac{3}{x-3}+\dfrac{5}{x-5}+\dfrac{17}{x-17}+\dfrac{19}{x-19}=x^2-11x-4

There are some positive integers a , b a,b and c 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

Rishabh Jain
Jan 4, 2016

3 x 3 + 1 + 5 x 5 + 1 + 17 x 17 + 1 + 19 x 19 + 1 = x 2 11 x \frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1=x^2-11x x ( 1 x 3 + 1 x 5 + 1 x 17 + 1 x 19 ) = x ( x 11 ) x(\color{#3D99F6}{\frac{1}{x-3}}+\color{#D61F06}{\frac{1}{x-5}+\frac{1}{x-17}}\color{#3D99F6}{+\frac{1}{x-19}})=x(x-11) 2 ( x 11 ) ( 1 ( x 3 ) ( x 19 ) + 1 ( x 5 ) ( x 17 ) ) = x 11 2(x-11)(\frac{1}{(x-3)(x-19)}+\frac{1}{(x-5)(x-17)})=x-11 Let x 2 22 x = t x = 11 ± t + 121 \color{#20A900}{\text{Let}\space x^{2}-22x=t \Rightarrow x=11\pm\sqrt{t+121}} 2 t + 57 + 2 t + 85 = 1 \Rightarrow \frac{2}{t+57}+\frac{2}{t+85}=1 t 2 + 138 t + 6885 = 0 t = 69 ± 200 \Rightarrow t^2+138t+6885=0 \Rightarrow t=-69\pm \sqrt{200} m = 11 + ( 69 + 200 ) + 121 \color{#69047E}{m=11+ \sqrt{ (-69+\sqrt{200})+121} } = 11 + ( 52 + 200 ) =11+\sqrt { (52+\sqrt{200} }) a + b + c = 11 + 52 + 200 = 263 a+b+c=11+52+200=\color{#3D99F6}{263}

Note that this is identical to Not the Brute Force Approach

Shaun Leong - 5 years, 5 months ago

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Why, oh, why, does Anish Harsha just LOVE repeating questions? (If only I knew this one was just repeated...)

Manuel Kahayon - 5 years, 5 months ago

My solution is a bit different in the sense that I have grouped appropriate terms to prevent a quartic equation(to simplify a bit) and at last I have not copied this solution from there(just saying), I am facing this question first time.

Rishabh Jain - 5 years, 5 months ago

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It is a great solution and similar to what I did when I first saw the problem. I was commenting on the repeated problem and did not want to post my comment as a solution - sorry if my comment seems to be referring to your solution.

Shaun Leong - 5 years, 5 months ago

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@Shaun Leong No no...it's all right.

Rishabh Jain - 5 years, 5 months ago

In fact, they're both copies of an AIME problem (too lazy to find it right now).

Daniel Liu - 5 years, 5 months ago
Aareyan Manzoor
Jan 5, 2016

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}

Thank you, @Aareyan Manzoor , I also like your solution . (+1)

Anish Harsha - 5 years, 5 months ago
Anupam Nayak
Jan 5, 2016

I just want to say that it would have been fun if it was similar to the original problem but not exactly the same.

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