Symbols are confusing

Algebra Level pending

φ − π 32 ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 32 + π ) = x \frac{\varphi -\pi^{32}}{(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)}= x If x x is given as above, where φ = 1 + 5 2 \varphi=\dfrac{1+\sqrt{5}}{2} is the golden ratio, then ( x + π ) 2016 = ( φ ) y (x+\pi)^{2016}= (\varphi)^y .

Find the value of y y .


The answer is 63.

Tommy Li by Tommy Li , 22, Hong Kong

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

x = φ − π 32 ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 32 + π ) = ( φ + π 16 ) ( φ − π 16 ) ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 32 + π ) = φ − π 16 ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 32 + π ) = φ 4 − π 8 ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 32 + π ) = φ 8 − π 4 ( φ 16 + π 2 ) ( φ 32 + π ) = φ 16 − π 2 φ 32 + π = φ 32 − π \begin{aligned} x & = \frac{\varphi -\pi^{32}}{(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)} \\ & = \frac{(\sqrt{\varphi} +\pi^{16}) (\sqrt{\varphi} - \pi^{16})} {(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)} \\ & = \frac{\sqrt{\varphi} - \pi^{16}} {(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)} \\ & = \frac{\sqrt[4]{\varphi} - \pi^{8}} {(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)} \\ & = \frac{\sqrt[8]{\varphi} - \pi^{4}} {(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)} \\ & = \frac{\sqrt[16]{\varphi} - \pi^{2}} {\sqrt[32]{\varphi}+\pi} \\ & =\sqrt[32]{\varphi}-\pi \end{aligned}

Now, we have:

( x + π ) 2016 = φ y ⟹ ( φ 32 − π + π ) 2016 = φ y ( φ 32 ) 2016 = φ y ⟹ y = 2016 32 = 63 \begin{aligned} (x+\pi) ^{2016} & =\varphi^y \\ \implies (\sqrt [32] {\varphi} - \pi+\pi) ^{2016} & =\varphi^y \\ (\sqrt [32] {\varphi} ) ^{2016} & =\varphi^y \\ \implies y&=\frac {2016}{32} \\&= \boxed {63} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for your solution ! We have quite different ways to solve the problem.

Tommy Li - 5 years ago
Tommy Li
Jun 6, 2016

= φ − π 32 ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 32 + π ) × ( φ 32 − π ) ( φ 32 − π ) \Large= \frac{\varphi -\pi^{32}}{(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[32]{\varphi}+\pi)} \times \frac{(\sqrt[32]{\varphi}-\pi)}{(\sqrt[32]{\varphi}-\pi)}

= φ − π 32 ( φ 32 − π ) ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 16 + π 2 ) ( φ 16 − π 2 ) \Large= \frac{\varphi -\pi^{32}(\sqrt[32]{\varphi}-\pi)}{(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[16]{\varphi}+\pi^2)(\sqrt[16]{\varphi}-\pi^2)}

= φ − π 32 ( φ 32 − π ) ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 8 + π 4 ) ( φ 8 − π 4 ) \Large= \frac{\varphi -\pi^{32}(\sqrt[32]{\varphi}-\pi)}{(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[8]{\varphi}+\pi^4)(\sqrt[8]{\varphi}-\pi^4)}

= φ − π 32 ( φ 32 − π ) ( φ + π 16 ) ( φ 4 + π 8 ) ( φ 4 − π 8 ) \Large= \frac{\varphi -\pi^{32}(\sqrt[32]{\varphi}-\pi)}{(\sqrt{\varphi}+\pi^{16})(\sqrt[4]{\varphi}+\pi^8)(\sqrt[4]{\varphi}-\pi^8)}

= φ − π 32 ( φ 32 − π ) ( φ + π 16 ) ( φ − π 16 ) \Large= \frac{\varphi -\pi^{32}(\sqrt[32]{\varphi}-\pi)}{(\sqrt{\varphi}+\pi^{16})(\sqrt{\varphi}-\pi^{16})}

= φ − π 32 ( φ 32 − π ) φ − π 32 \Large= \frac{\varphi -\pi^{32}(\sqrt[32]{\varphi}-\pi)}{\varphi -\pi^{32}}

= φ 32 − π =\sqrt[32]{\varphi}-\pi

( ( φ 32 − π ) + π ) 2016 ((\sqrt[32]{\varphi}-\pi)+\pi)^{2016}

= φ 32 2016 =\sqrt[32]{\varphi}^{2016}

= φ 2016 32 =\varphi^{\frac{2016}{32}}

= φ 63 =\varphi^{63}

y = 63 y=63

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...