Maths Olympiad Prep

Library / Estonian Math Competitions /59 of 148

Estonian Math Competitions

Number theory Difficulty 5.5 AIME, harder Prove it Estonia

Find the least positive integer nn such that 5n5\sqrt[5]{5n}, 6n6\sqrt[6]{6n} and 7n7\sqrt[7]{7n} are integers.

Solution

Let n=2α⋅3β⋅5γ⋅7δ⋅sn = 2^\alpha \cdot 3^\beta \cdot 5^\gamma \cdot 7^\delta \cdot s, where ss is not divisible by 22, 33, 55 or 77; then 5n=2α⋅3β⋅5γ+1⋅7δ⋅s5n = 2^\alpha \cdot 3^\beta \cdot 5^{\gamma+1} \cdot 7^\delta \cdot s, 6n=2α+1⋅3β+1⋅5γ⋅7δ⋅s6n = 2^{\alpha+1} \cdot 3^{\beta+1} \cdot 5^\gamma \cdot 7^\delta \cdot s and 7n=2α⋅3β⋅5γ⋅7δ+1⋅s7n = 2^\alpha \cdot 3^\beta \cdot 5^\gamma \cdot 7^{\delta+1} \cdot s. Consequently:
* For 5n5\sqrt[5]{5n} to be an integer, α\alpha, β\beta, γ+1\gamma + 1 and δ\delta must be divisible by 55;
* For 6n6\sqrt[6]{6n} to be an integer, α+1\alpha + 1, β+1\beta + 1, γ\gamma and δ\delta must be divisible by 66;
* For 7n7\sqrt[7]{7n} to be an integer, α\alpha, β\beta, γ\gamma and δ+1\delta + 1 must be divisible by 77.
Hence α\alpha and β\beta must be divisible by 3535, γ\gamma must be divisible by 4242 and δ\delta must be divisible by 3030. The least suitable value for α\alpha and β\beta is 3535 since 3535 is the least positive multiple of 3535 and the next integer is divisible by 66. Studying the multiples of 4242 and 3030 similarly shows that the least suitable value for γ\gamma is 8484 and the least suitable value for δ\delta is 9090. For the least suitable value for nn, take s=1s = 1. Hence the desired number is 235⋅335⋅584⋅7902^{35} \cdot 3^{35} \cdot 5^{84} \cdot 7^{90}.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.