MathDB
Prove that $2\log_3 n \leq f(n) < 5\log_3 n$ for every $n>1$.

Source: Moldova TST 2001

August 6, 2023

Problem Statement

For every nonnegative integer nn{} let f(n)f(n) be the smallest number of digits 11 which can represent the number nn{} using the symbols "+","","×","(",")""+", "-", "\times", "(", ")". For example, 80=(1+1+1+1+1)×(1+1+1+1)×(1+1+1+1)80=(1+1+1+1+1)\times(1+1+1+1)\times(1+1+1+1) and f(80)13f(80)\leq 13. Prove that 2log3nf(n)<5log3n2\log_3 n \leq f(n) < 5\log_3 n for every n>1n>1.