Find-the-100th-digits-of-this-sum-1-50-50-2-50-999-

Question Number 6831 by Tawakalitu. last updated on 30/Jul/16

F i n d t h e 100 t h d i g i t s o f t h i s s u m

1 + 50 + 50^{2} + \dots . . + 50^{999}

Commented by prakash jain last updated on 31/Jul/16

How did you get

5^{1000} \times 10^{1000} - 1 = (5^{1000} - 1) \times (999 \dots 9999) ?

Commented by Yozzii last updated on 30/Jul/16

\underset{i = 0}{\sum^{999}} 50^{i} = \frac{50^{1000} - 1}{50 - 1} = \frac{50^{1000} - 1}{49}

= \frac{5^{1000} \times 10^{1000} - 1}{49}

N o w, n \times 10^{m} = n 000 \dots 000 (m z e r o e s) (n \in N)

\Rightarrow n 10^{m} - 1 = (n - 1) 999 \dots 999 (m n i n e s)

∴ 5^{1000} \times 10^{1000} = (5^{1000}) 000 \dots 000 (1000 z e r o e s t r a i l i n g)

\Rightarrow 5^{1000} \times 10^{1000} - 1 = (5^{1000} - 1) 999 \dots 999 (1000 n i n e s t r a i l i n g)

(c o n t i n u e)

Commented by Yozzii last updated on 31/Jul/16

I w a s t r y i n g t o w r i t e t h e n u m b e r

i n d e c i m a l f o r m .

e . g 5^{2} \times 10^{3} = 25000

\Rightarrow 5^{2} \times 10^{3} - 1 = 24999 = (5^{2} - 1) 999 (i n d e c i m a l

r e p r e s e n t a t i o n, n o t a p r o d u c t)

Commented by Tawakalitu. last updated on 31/Jul/16

T h a n k s s i r

Commented by Tawakalitu. last updated on 31/Jul/16

T h a n k s s i r