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) ) )))))


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

$define d (n) to be the sum of the digits$ $of n .$ $i . e d (1000) = 1, d (999) = 27$ $find d (d (d (d (d (5^{10^{100}})))))$
Commented by Rasheed.Sindhi last updated on 08/Sep/21

$3 ∤ d (d (d (d (d (5^{10^{100}})))))$ $I - E T h e r e q u i r e d n u m b e r i s n o t$ $m u l t i p l e o f 3$
Commented by talminator2856791 last updated on 08/Sep/21

$proof ?$
Commented by Rasheed.Sindhi last updated on 09/Sep/21

$I f y o u s u f f i c i e n t l y a p p l y d t o g e t$ $a n s w e r i n o n e d i g i t y o u w i l l u l t i m a t e l y$ $g e t 4 . I f y o u w a n t o n e d i g i t a n s w e r$ $I h a v e a s o l u t i o n .$
Commented by talminator2856791 last updated on 09/Sep/21

$wheres the proof ?$
	Answered by Rasheed.Sindhi last updated on 10/Sep/21
	$Relation between sum of digits and remainder modulo 9: 5^(10^(100) ) ≡d(d(d(d(d(5^(10^(100) ) )))))(mod 9) We can verify easily that 5^6 ≡1(mod 9) (5^6 )^k ≡1^k (mod 9) 5^(6k) ≡1(mod 9) { ((5^(6k) ≡1(mod 9))),((5^(6k+1) ≡5(mod 9))),((5^(6k+2) ≡7(mod 9))),((5^(6k+3) ≡8(mod 9))),((5^(6k+4) ≡4(mod 9)^★ )),((5^(6k+5) ≡2(mod 9))) :} 5^(10^(100) ) =5^(6k+ ?) { ((10≡4(mod 6))),((10^2 ≡4(mod 6)),((...)),((10^(100) ≡4(mod 6))) :} 5^(10^(100) ) =5^(6k+ 4) d(d(d(d(d(5^(10^(100) ) )))))≡5^(10^(100) ) (mod 9) d(d(d(d(d(5^(10^(100) ) )))))≡5^(6k+4) (mod 9) But we′ve determined before that 5^(6k+4) ≡4(mod 9)^★ ∴d(d(d(d(d(5^(10^(100) ) )))))≡4(mod 9) ∴ Ultimately sum of digits will be 4 if we apply sufficiently the function d over the number 5^(10^(100) ) .$
	$R e l a t i o n b e t w e e n sum of digits a n d$ $remainder modulo 9 :$ $5^{10^{100}} \equiv d (d (d (d (d (5^{10^{100}}))))) (m o d 9)$ $W e c a n v e r i f y e a s i l y t h a t$ $5^{6} \equiv 1 (m o d 9)$ ${(5^{6})}^{k} \equiv 1^{k} (m o d 9)$ $5^{6 k} \equiv 1 (m o d 9)$ ${\begin{cases} 5^{6 k} \equiv 1 (m o d 9) \\ 5^{6 k + 1} \equiv 5 (m o d 9) \\ 5^{6 k + 2} \equiv 7 (m o d 9) \\ 5^{6 k + 3} \equiv 8 (m o d 9) \\ 5^{6 k + 4} \equiv 4 {(m o d 9)}^{★} \\ 5^{6 k + 5} \equiv 2 (m o d 9) \end{cases}$ $5^{10^{100}} = 5^{6 k + ?}$ ${\begin{cases} 10 \equiv 4 (m o d 6) \\ 10^{2} \equiv 4 (m o d 6 \\ . . . \\ 10^{100} \equiv 4 (m o d 6) \end{cases}$ $5^{10^{100}} = 5^{6 k + 4}$ $d (d (d (d (d (5^{10^{100}}))))) \equiv 5^{10^{100}} (m o d 9)$ $d (d (d (d (d (5^{10^{100}}))))) \equiv 5^{6 k + 4} (m o d 9)$ $B u t {w e}^{'} v e d e t e r m i n e d b e f o r e t h a t$ $5^{6 k + 4} \equiv 4 {(m o d 9)}^{★}$ $∴ d (d (d (d (d (5^{10^{100}}))))) \equiv 4 (m o d 9)$ $∴ U l t i m a t e l y s u m o f d i g i t s w i l l b e$ $4 i f w e a p p l y s u f f i c i e n t l y t h e f u n c t i o n$ $d o v e r t h e n u m b e r 5^{10^{100}} .$
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