Math is Crazy

Algebra Level 2

Given that x , y x,y and z z are integers, if 50 13 = 3 + 1 x + 1 y + 1 z \frac{50}{13} = 3 + \frac{1}{x+\frac{1}{y+\frac{1}{z}}} , find x + y + z x+y+z .


The answer is 8.

Jason Chrysoprase by Jason Chrysoprase , 18, Indonesia

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1 solution

Jason Chrysoprase
Jan 30, 2016

Let us make 50 13 \frac{50}{13} in the form of 3 + 1 x + 1 y + 1 z 3 + \frac{1}{x+\frac{1}{y+\frac{1}{z}}} ,

50 13 \frac{50}{13} = 3 + 11 13 = 3 + \frac{11}{13} = 3 + 1 13 11 =3+\frac{1}{\frac{13}{11}} = 3 + 1 1 + 2 11 =3+\frac{1}{1 +\frac{2}{11}} = 3 + 1 1 + 1 11 2 =3+\frac{1}{1 +\frac{1}{\frac{11}{2}}} = 3 + 1 1 + 1 5 + 1 2 =3+\frac{1}{1 +\frac{1}{5+\frac{1}{2}}}

Therefore, x = 1 x = 1 , y = 5 y=5 and z = 2 z=2

So, x + y + z = 1 + 5 + 2 = 8 x+y+z= 1+5+2= 8

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Z will be 4 I can prove it

Rishav Agarwal - 4 years, 6 months ago

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Sure thing, but when we sum x,y,z , will it still be 8 ?

Jason Chrysoprase - 4 years, 6 months ago

If you recommend then I will post the answer

Rishav Agarwal - 4 years, 6 months ago

You should mention x,y,z are integers. For example, there is a solution (0, 10/13, 13), but the sum is 179/13 which is not correct from your point of view.

Qantas Perth - 4 years, 1 month ago

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Ahh yes, i'm sorry i didn't notice that, it's my old problem :p

Jason Chrysoprase - 4 years, 1 month ago

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