Let m be the the largest real solution to the equation
x − 3 3 + x − 5 5 + x − 1 7 1 7 + x − 1 9 1 9 = x 2 − 1 1 x − 4
There are some positive integers a , b and c such that m = a + b + c . Find a + b + c .
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Note that this is identical to Not the Brute Force Approach
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Why, oh, why, does Anish Harsha just LOVE repeating questions? (If only I knew this one was just repeated...)
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.
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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.
In fact, they're both copies of an AIME problem (too lazy to find it right now).
first off.. i loved this problem, thanks to the creator for giving me such great experience. start like this x − 3 3 + x − 5 5 + x − 1 7 1 7 + x − 1 9 1 9 + 4 = x 2 − 1 1 x x − 3 3 + 1 + x − 5 5 + 1 + x − 1 7 1 7 + 1 + x − 1 9 1 9 + 1 = x 2 − 1 1 x x − 3 x + x − 5 x + x − 1 7 x + x − 1 9 x = x ( x − 1 1 ) we get x=0 as a solution. now divide by x and put x=y+11 y + 8 1 + y − 8 1 + y + 6 1 + y − 6 1 = y y 2 − 6 4 2 y + y 2 − 3 6 2 y = y y=0 ->x=11 is another solution. we divide again to get y 2 − 6 4 2 + y 2 − 3 6 2 = 1 2 ( y 2 − 3 6 ) + 2 ( y 2 − 6 4 ) = ( y 2 − 6 4 ) ( y 2 − 3 6 ) 4 y 2 − 2 0 0 = y 4 − 1 0 0 y 2 + 2 3 0 4 y 4 − 1 0 4 y 2 + 2 5 0 4 = 0 y 2 = 5 2 ± 2 0 0 y = ± 5 2 ± 2 0 0 x = 1 1 ± 5 2 ± 2 0 0 we get that x = 0 , 1 1 , 1 1 ± 5 2 ± 2 0 0 the required x l a r g e s t = 1 1 + 5 2 + 2 0 0 1 1 + 5 2 + 2 0 0 = 2 6 3
Thank you, @Aareyan Manzoor , I also like your solution . (+1)
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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x − 3 3 + 1 + x − 5 5 + 1 + x − 1 7 1 7 + 1 + x − 1 9 1 9 + 1 = x 2 − 1 1 x x ( x − 3 1 + x − 5 1 + x − 1 7 1 + x − 1 9 1 ) = x ( x − 1 1 ) 2 ( x − 1 1 ) ( ( x − 3 ) ( x − 1 9 ) 1 + ( x − 5 ) ( x − 1 7 ) 1 ) = x − 1 1 Let x 2 − 2 2 x = t ⇒ x = 1 1 ± t + 1 2 1 ⇒ t + 5 7 2 + t + 8 5 2 = 1 ⇒ t 2 + 1 3 8 t + 6 8 8 5 = 0 ⇒ t = − 6 9 ± 2 0 0 m = 1 1 + ( − 6 9 + 2 0 0 ) + 1 2 1 = 1 1 + ( 5 2 + 2 0 0 ) a + b + c = 1 1 + 5 2 + 2 0 0 = 2 6 3