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Question Number 153565 by talminator2856791 last updated on 08/Sep/21

define d(n) to be the sum of the digits of n. i.e d(1000) = 1 , d(999) = 27 find d(d(d(d(d(5^(10^(100) ) )))))

$$\: \\ $$$$\:\mathrm{define}\:{d}\left({n}\right)\:\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\: \\ $$$$\:\mathrm{of}\:{n}.\:\:\: \\ $$$$\:\mathrm{i}.\mathrm{e}\:\:{d}\left(\mathrm{1000}\right)\:=\:\mathrm{1}\:,\:\:\:\:{d}\left(\mathrm{999}\right)\:=\:\mathrm{27} \\ $$$$\: \\ $$$$\:\mathrm{find}\:\:{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right) \\ $$$$\: \\ $$

Commented by Rasheed.Sindhi last updated on 08/Sep/21

3 ∤ d(d(d(d(d(5^(10^(100) ) ))))) I-E The required number is not multiple of 3

$$\mathrm{3}\:\nmid\:{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right) \\ $$$$\mathrm{I}-\mathrm{E}\:\mathcal{T}{he}\:{required}\:{number}\:{is}\:{not} \\ $$$${multiple}\:{of}\:\mathrm{3} \\ $$

Commented by talminator2856791 last updated on 08/Sep/21

proof?

$$\:\mathrm{proof}? \\ $$

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 4. If you want one digit answer I have a solution.

$${If}\:{you}\:{sufficiently}\:{apply}\:{d}\:{to}\:{get} \\ $$$${answer}\:{in}\:{one}\:{digit}\:{you}\:{will}\:{ultimately} \\ $$$${get}\:\mathrm{4}.\:{If}\:{you}\:{want}\:{one}\:{digit}\:{answer} \\ $$$${I}\:{have}\:{a}\:{solution}. \\ $$

Commented by talminator2856791 last updated on 09/Sep/21

wheres the proof?

$$\:\mathrm{wheres}\:\mathrm{the}\:\mathrm{proof}? \\ $$

Answered by Rasheed.Sindhi last updated on 10/Sep/21

Relation between sum of digits and remainder modulo 9: 5^(10^(100) ) ≡d(d(d(d(d(5^(10^(100) ) )))))(mod 9) We can verify easily that 5^6 ≡1(mod 9) (5^6 )^k ≡1^k (mod 9) 5^(6k) ≡1(mod 9) { ((5^(6k) ≡1(mod 9))),((5^(6k+1) ≡5(mod 9))),((5^(6k+2) ≡7(mod 9))),((5^(6k+3) ≡8(mod 9))),((5^(6k+4) ≡4(mod 9)^★ )),((5^(6k+5) ≡2(mod 9))) :} 5^(10^(100) ) =5^(6k+ ?) { ((10≡4(mod 6))),((10^2 ≡4(mod 6)),((...)),((10^(100) ≡4(mod 6))) :} 5^(10^(100) ) =5^(6k+ 4) d(d(d(d(d(5^(10^(100) ) )))))≡5^(10^(100) ) (mod 9) d(d(d(d(d(5^(10^(100) ) )))))≡5^(6k+4) (mod 9) But we′ve determined before that 5^(6k+4) ≡4(mod 9)^★ ∴d(d(d(d(d(5^(10^(100) ) )))))≡4(mod 9) ∴ Ultimately sum of digits will be 4 if we apply sufficiently the function d over the number 5^(10^(100) ) .

$$ \\ $$$$\mathcal{R}{elation}\:{between}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{digits}\:{and} \\ $$$$\mathrm{remainder}\:\mathrm{modulo}\:\mathrm{9}: \\ $$$$\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \equiv{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\left({mod}\:\mathrm{9}\right) \\ $$$${We}\:{can}\:{verify}\:{easily}\:{that} \\ $$$$\mathrm{5}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\left(\mathrm{5}^{\mathrm{6}} \right)^{{k}} \equiv\mathrm{1}^{{k}} \left({mod}\:\mathrm{9}\right) \\ $$$$\mathrm{5}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\begin{cases}{\mathrm{5}^{\mathrm{6}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{1}} \equiv\mathrm{5}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{2}} \equiv\mathrm{7}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{3}} \equiv\mathrm{8}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{4}} \equiv\mathrm{4}\left({mod}\:\mathrm{9}\right)^{\bigstar} }\\{\mathrm{5}^{\mathrm{6}{k}+\mathrm{5}} \equiv\mathrm{2}\left({mod}\:\mathrm{9}\right)}\end{cases} \\ $$$$\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } =\mathrm{5}^{\mathrm{6}{k}+\:?} \\ $$$$\begin{cases}{\mathrm{10}\equiv\mathrm{4}\left({mod}\:\mathrm{6}\right)}\\{\mathrm{10}^{\mathrm{2}} \equiv\mathrm{4}\left({mod}\:\mathrm{6}\right.}\\{...}\\{\mathrm{10}^{\mathrm{100}} \equiv\mathrm{4}\left({mod}\:\mathrm{6}\right)}\end{cases} \\ $$$$\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } =\mathrm{5}^{\mathrm{6}{k}+\:\mathrm{4}} \\ $$$${d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\equiv\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \left({mod}\:\mathrm{9}\right) \\ $$$${d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\equiv\mathrm{5}^{\mathrm{6}{k}+\mathrm{4}} \left({mod}\:\mathrm{9}\right) \\ $$$${But}\:{we}'{ve}\:{determined}\:{before}\:{that} \\ $$$$\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{6}{k}+\mathrm{4}} \equiv\mathrm{4}\left({mod}\:\mathrm{9}\right)^{\bigstar} \\ $$$$\therefore{d}\left({d}\left({d}\left({d}\left({d}\left(\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \right)\right)\right)\right)\right)\equiv\mathrm{4}\left({mod}\:\mathrm{9}\right) \\ $$$$\therefore\:{Ultimately}\:{sum}\:{of}\:{digits}\:{will}\:{be} \\ $$$$\mathrm{4}\:{if}\:{we}\:{apply}\:{sufficiently}\:{the}\:{function} \\ $$$$\boldsymbol{{d}}\:{over}\:{the}\:{number}\:\:\mathrm{5}^{\mathrm{10}^{\mathrm{100}} } \:. \\ $$$$ \\ $$

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