Is-there-a-formular-to-tell-how-many-times-a-digit-occur-in-an-interval-e-g-How-many-times-digits-2-occur-between-1-100-

Question Number 116091 by I want to learn more last updated on 30/Sep/20

Is there a formular to tell how many times a digit occur in an interval .

e . g . How many times digits 2 occur between 1 - 100

Answered by 1549442205PVT last updated on 01/Oct/20

In each tens the digits 2 appears one

time, only tens from 20 to 30 digits 2

appears two times . Therefore, from 1

to 100 digts 2 appears 11 times

The other digits are the same

Commented by I want to learn more last updated on 01/Oct/20

Thanks sir

Commented by 1549442205PVT last updated on 01/Oct/20

You are welcome

Commented by mr W last updated on 02/Oct/20

2 \Rightarrow o n e t i m e 2

12, 22, 32, \dots, 92 \Rightarrow 10 t i m e s 2

20, 21, 23, \dots, 29 \Rightarrow 9 t i m e s 2

\Rightarrow t o t a l l y 1 + 10 + 9 = 20 t i m e s 2

Answered by mr W last updated on 01/Oct/20

n d i g i t n u m b e r s :

10^{n - 1} + (n - 1) \times 9 \times 10^{n - 2}

= (9 n + 1) 10^{n - 2}

f r o m 1 t o 100 :

1 + 19 \times 1 = 20 t i m e s

Commented by I want to learn more last updated on 06/Oct/20

I just see this now sir, so it will work for 1 to 50 too or

another nth derivative

Commented by I want to learn more last updated on 06/Oct/20

And sorry for bordering you sir,

How to generalise it as you solve, so that i can think too on my own .

Thanks sir .

Commented by I want to learn more last updated on 06/Oct/20

Thanks sir, i appreciate .

Commented by mr W last updated on 06/Oct/20

{l e t}^{'} s l o o k a t n u m b e r s w i t h m d i g i t s

x x x x \dots x

t h e d i g i t 2 (o r a n y o t h e r n o n - z e r o

d i g i t) c a n b e a t f i r s t p o s i t i o n

2 x x x . . x

t h e r e a r e 10 \times 10 \times . . \times 10 = 10^{m - 1}

s u c h n u m b e r s, i . e . d i g i t 2 c o m e s

10^{m - 1} t i m e s a t t h i s p o s i t i o n .

w h e n t h e d i g i t 2 i s a t o t h e r p o s i t i o n s,

x 2 x x . . x (m - 1 p o s s i b l e p o s i t i o n s)

t h e r e a r e 9 \times 10 \times 10 \times \dots 10 = 9 \times 10^{m - 2}

s u c h n u m b e r s, i . e . d i g i t 2 c o m e s

(m - 1) \times 9 \times 10^{m - 2} t i m e s .

a l l t o g e t h e r :

10^{m - 1} + (m - 1) \times 9 \times 10^{m - 2}

= 10^{m - 2} [10 + (m - 1) \times 9]

= (9 m + 1) \times 10^{m - 2}

Commented by I want to learn more last updated on 06/Oct/20

I appreciate sir