# UKMT Special (Problem $13$)

For each positive real integer $x$, we define {$x$} to be the greater of $\frac{1}{x}$ and $x$, with {$1$} $=1$

Find, with proof, all positive real numbers $y$ such that

$5y \{ 8y \} \{ 25y \} = 1$

[UKMT BMO $2016$ Round $1$, Q$2$]

Note by Yajat Shamji
8 months, 2 weeks ago

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You have until next Thursday, $3:00$pm!

- 8 months, 2 weeks ago

Step 1) {x} is the greater of 1/x and x, with {1} = 1

   If x >= 1, we can divide both by x and get 1 >= 1/x. This implies that x >= 1 >= 1/x or x >= 1/x. This suggests that {x} = x.
If 1 > x >= 0, we can divide both by x and get 1/x > 1 >= 0 or 1/x > 1. This implies that 1/x > 1 > x >= 0 or 1/x > x. This suggests that {x} = 1/x.


I find the solutions by dividing to 3 case. The solution is y = 1/40 or y = 8/125. But somehow, all my work that I typed disappear. I'm not writing it again.

- 8 months, 2 weeks ago

- 8 months, 2 weeks ago

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