MathDB

5

Part of 2015 Czech-Polish-Slovak Junior Match

Problems(2)

For each divisor d> 1 of n, then (d - 1) is a divisor of (n - 1|)

Source: Czech-Polish-Slovak Junior Match 2015, Individual p5 CPSJ

3/15/2020
Determine all natural numbersn>1 n> 1 with the property: For each divisor d>1d> 1 of number nn, then d−1d - 1 is a divisor of n−1n - 1.
divisornumber theory
\sqrt{ab}- 2ab/(a + b) \le p ( (a + b)/2} -\sqrt{ab} ) , in p for any a,b>0

Source: Czech-Polish-Slovak Junior Match 2015, Team p5 CPSJ

3/19/2020
Find the smallest real constant pp for which the inequality holds ab−2aba+b≤p(a+b2−ab)\sqrt{ab}- \frac{2ab}{a + b} \le p \left( \frac{a + b}{2} -\sqrt{ab}\right) with any positive real numbers a,ba, b.
inequalitiesminimumminalgebra