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-

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

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 \\ \dots \\ 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}} .