olymmath

IMC 2023 Problem 10

    For every positive integer nn , let f(n)f(n) , g(n)g(n) be the minimal positive integers such that 1+11!+12!+⋯+1n!=f(n)g(n).1+\frac{1}{1!}+\frac{1}{2!}+\dots +\frac{1}{n!}=\frac{f(n)}{g(n)}. Determine whether there exists a positive integer nn for which g(n)>n0.999ng(n)>n^{0.999n} .

    Dataset text AI-MO AoPS. Quoted with source.

    Official text: International Mathematics Competition for University Students 2023 official paper, page 1.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator imc/2023/p10 · Tag F3QZDV4CDE · Edit this page on GitHub · Keys: j next, k previous