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

2 Responses to “Le Monde puzzle [47]”

  1. Jean Louis FOULLEY Says:

    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*

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