N

Hi @jollyorangutan,
Nice question! It is wonderful to be curious and think about the solutions over and over again in order to get to the bottom of it!
In this problem, you can easily check your assumption. The numbers 0.5487, 0.5492 and 0.5502 are so small that we can count them as equal in the beginning. After multiplying all of these three numbers (that we assumed are equal), we get a number \(^1/_n,\) where \(n\) is an integer. You think that \(n=7\) will be the closer number than \(n=6\) (Btw, why not \(n=5\)? All numbers (0.5487, 0.5492 and 0.5502) were smaller than 0.555..., so why did you choose 7 instead of 5? ). Let's check which one is closer!
$$0.5487\times 0.5492\times 0.5502=({\color{blue}\overline{0.---...}})^3=\frac{1}{\color{purple}\bf{n}}$$
$${\color{blue}\overline{0.---...}}=\sqrt[\bf{3}]{\frac{1}{\color{purple}\bf{n}}}$$
where \({\color{blue}\overline{0.---...}}\) is the number that is close to each one of 0.5487, 0.5492 and 0.5502, so that the number \(0.5487\times 0.5492\times 0.5502\) could be interpreted as equal to \(({\color{blue}\overline{0.---...}})^3.\)
So, we know that \(\sqrt[\bf{3}]{\frac{1}{\color{purple}\bf{n}}}={\color{blue}\overline{0.---...}}\). Then let's check our \(n=6\) and \(n=7\) (if you want, you can check \(n=5\) later by yourself):
$${\color{blue}\overline{0.---...}}=\sqrt[\bf{3}]{\frac{1}{\color{purple}\bf{6}}}=0.550321...\approx {\color{purple}0.5503}\text{ - well, pretty close, don't you think?}$$ $${\color{blue}\overline{0.---...}}=\sqrt[\bf{3}]{\frac{1}{\color{purple}\bf{7}}}=0.522757...\approx {\color{purple}0.5228}\text{ - hm, I think the first one was closer)}$$
As for me, the number 0.5228 is much further than 0.5503 from our numbers 0.5487, 0.5492 and 0.5502, so \(n=6\) is closer than \(n=7\)
(What do you think about \(\it{n=5?}\))