Le Monde puzzle [47]

The weekend Le Monde puzzle sounds [once again] too easy:

If y is the integer part of the positive (non-negative) real number x and z=x-y, find all x‘s such that there exists a factor a with x=ay and y=az.

Given that x=y+z, we must have a²z=(a+1)z, which leads to the unique factor

$a^*=\dfrac{1+\sqrt{5}}{2},$

which is less than 2. Furthermore, since y is an integer, z=y/a^* with y<a^*<2. This restricts the choice to y=0, leading to x=0 and y=1, leading to x=a…

This entry was posted on November 24, 2010 at 12:47 am and is filed under Statistics with tags Le Monde, mathematical puzzle. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Le Monde puzzle [47]”

Jean Louis FOULLEY Says:
November 24, 2010 at 11:48 pm

If my memory serves me right: the ratio (y+z)/y=z/y is by definition the famous (and ubiquitous) golden ratio “phi” ie your a*

Reply
- xi'an Says:
  November 25, 2010 at 5:00 am
  
  Thank you, Jean-Louis!, I was indeed thinking that this number
  
  $\varphi=\dfrac{1+\sqrt{5}}{2}$
  
  was sounding like the golden ratio but I didn’t check!</p.
  
  Reply

Xi'an's Og