Question Number 153565 by talminator2856791 last updated on 08/Sep/21
$$\: \\ $$$$\:\mathrm{define}\:{d}\left({n}\right)\:\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\: \\ $$$$\:\mathrm{of}\:{n}.\:\:\: \\ $$$$\:\mathrm{i}.\mathrm{e}\:\:{d}\left(\mathrm{1000}\right)\:=\:\mathrm{1}\:,\:\:\:\:{d}\left(\mathrm{999}\right)\:=\:\mathrm{27} \\ $$$$\: \\ $$$$\:\mathrm{find}\:\:{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right) \\ $$$$\: \\ $$
Commented by Rasheed.Sindhi last updated on 08/Sep/21
$$\mathrm{3}\:\nmid\:{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right) \\ $$$$\mathrm{I}-\mathrm{E}\:\mathcal{T}{he}\:{required}\:{number}\:{is}\:{not} \\ $$$${multiple}\:{of}\:\mathrm{3} \\ $$
Commented by talminator2856791 last updated on 08/Sep/21
$$\:\mathrm{proof}? \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/21
$${If}\:{you}\:{sufficiently}\:{apply}\:{d}\:{to}\:{get} \\ $$$${answer}\:{in}\:{one}\:{digit}\:{you}\:{will}\:{ultimately} \\ $$$${get}\:\mathrm{4}.\:{If}\:{you}\:{want}\:{one}\:{digit}\:{answer} \\ $$$${I}\:{have}\:{a}\:{solution}. \\ $$
Commented by talminator2856791 last updated on 09/Sep/21
$$\:\mathrm{wheres}\:\mathrm{the}\:\mathrm{proof}? \\ $$
Answered by Rasheed.Sindhi last updated on 10/Sep/21
$$ \\ $$$$\mathcal{R}{elation}\:{between}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{digits}\:{and} \\ $$$$\mathrm{remainder}\:\mathrm{modulo}\:\mathrm{9}: \\ $$$$\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \equiv{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\left({mod}\:\mathrm{9}\right) \\ $$$${We}\:{can}\:{verify}\:{easily}\:{that} \\ $$$$\mathrm{5}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\left(\mathrm{5}^{\mathrm{6}} \right)^{{k}} \equiv\mathrm{1}^{{k}} \left({mod}\:\mathrm{9}\right) \\ $$$$\mathrm{5}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\begin{cases}{\mathrm{5}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{1}} \equiv\mathrm{5}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{2}} \equiv\mathrm{7}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{3}} \equiv\mathrm{8}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{4}} \equiv\mathrm{4}\left({mod}\:\mathrm{9}\right)^{\bigstar} }\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{5}} \equiv\mathrm{2}\left({mod}\:\mathrm{9}\right)}\end{cases} \\ $$$$\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } =\mathrm{5}^{\mathrm{6}{k}+\:?} \\ $$$$\begin{cases}{\mathrm{10}\equiv\mathrm{4}\left({mod}\:\mathrm{6}\right)}\\{\mathrm{10}^{\mathrm{2}} \equiv\mathrm{4}\left({mod}\:\mathrm{6}\right.}\\{…}\\{\mathrm{10}^{\mathrm{100}} \equiv\mathrm{4}\left({mod}\:\mathrm{6}\right)}\end{cases} \\ $$$$\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } =\mathrm{5}^{\mathrm{6}{k}+\:\mathrm{4}} \\ $$$${d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\equiv\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \left({mod}\:\mathrm{9}\right) \\ $$$${d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\equiv\mathrm{5}^{\mathrm{6}{k}+\mathrm{4}} \left({mod}\:\mathrm{9}\right) \\ $$$${But}\:{we}'{ve}\:{determined}\:{before}\:{that} \\ $$$$\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{6}{k}+\mathrm{4}} \equiv\mathrm{4}\left({mod}\:\mathrm{9}\right)^{\bigstar} \\ $$$$\therefore{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\equiv\mathrm{4}\left({mod}\:\mathrm{9}\right) \\ $$$$\therefore\:{Ultimately}\:{sum}\:{of}\:{digits}\:{will}\:{be} \\ $$$$\mathrm{4}\:{if}\:{we}\:{apply}\:{sufficiently}\:{the}\:{function} \\ $$$$\boldsymbol{{d}}\:{over}\:{the}\:{number}\:\:\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \:. \\ $$$$ \\ $$