How many times is digit 0 written when listing all numbers from 1 to 3333?


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Question Number 18498 by Tinkutara last updated on 22/Jul/17

$How many times is digit 0 written when$ $listing all numbers from 1 to 3333 ?$
Commented by richard last updated on 23/Jul/17

$d o y o u w a n t h o w m a n y z e r o s t h e r e a r e b e t w e e n 1 a n d 3333, o r h o w m a n y t i m e s t h e ‘ ‘ 0^{″} a p p e a r s, i n d e p e n d e n t l y o f t h e a m o u n t o f z e r o s ?$
	Answered by Abbas-Nahi last updated on 23/Jul/17

	$I t h i n t h e t i m e s o f d i g i t i s$ $9 \times 10 \times 10 \times 1 = 900$
	Answered by richard last updated on 23/Jul/17

	$f r o m 1 t o 100 \to 10 t i m e s (10, 20, 30 . . . 100)$ $f r o m 101 t o 200 \to 19 t i m e s (10 t i m e s t h e z e r o (o r z e r o s a p p e a r s) f r o m 101 t o 110 p l u s 9 t i m e s t h e z e r o a p p e a r s f r o m 111 t o 1000)$ $c o n s i d e r i n g t h i s p a d r o n f o r t h e h u n d r e d s t o t h e t h o u s a n d s,$ $\to 101 t o 1000 \equiv 19 \cdot 9 = 181 t i m e s$ $f o r t h e t h o u s a n d s, w e c o u l d s a y t h a t f o r 1001 t o 1100 = 100 t i m e s (t h e z e r o o r m o r e t h a n o n e z e r o a p p e a r s)$ $A s b e f o r e, f r o m 1100 t o 2000 = 19 \cdot 9 = 181$ $t h i s w a y, k n o w i n g t h i s p a d r o n f o r$ $d o z e n s, h u n d r e d s a n d t h o u s a n d s :$ $2000 \to 2100 = 100 t i m e s$ $2101 \to 3000 = 181 t i m e s$ $3001 \to 3100 = 100 t i m e s$ $3101 \to 3300 = 19 \cdot 2 = 38 t i m e s$ $3301 \to 3333 = 10 + 2 = 12 t i m e s$ $t o t a l :$ $Σ = 10 + 181 + 100 + 181 + 100 + 181 + 100 + 38 + 12$ $Σ = 100 \cdot 3 + 181 \cdot 3 + 10 + 38 + 12$ $Σ = 300 + 543 + 60$ $Σ = 903 t i m e s (o n c e m o r e, w h e n i s a y ‘ ‘ {t i m e s}^{″}, i m e a n t h e n u m b e r o f t i m e s t h e z e r o o r m o r e t h a n o n e z e r o a p p e a r s b e t w e e n 1 a n d 3333)$
	Commented by mrW1 last updated on 23/Jul/17

	$f r o m 1 t o 100 \to 11 z e r o s (10, 20, 30 . . . 100)$
	Commented by richard last updated on 23/Jul/17

	$w h y 11 ? t h e q u e s t i o n a s k s a b o u t h o w m a n y t i m e s t h e z e r o a p p e a r s, n o t h o w m a n y z e r o s e x i s t s b e t w e e n t h e n, i s i t n o t ?$
	Commented by richard last updated on 23/Jul/17

	$I n e e d t o f i x i t, t h d c o r r e c t a n s w e r i s$ $Σ = 903 t i m e s$ $I t m e a n s t h e z e r o (o r m o r e z e r o s) r a p p e a r s 903 t i m e s b e t w e e n 1 a n d 3333, n o t h o w m a n y z e r o s a r e b e t w e e n 1 a n d 3333$
	Commented by richard last updated on 23/Jul/17

	$t h e *$
	Commented by richard last updated on 23/Jul/17

	$I t s h o u l d b e Σ = 903 t i m e s, n o t Σ = 903 z e r o s, s o r r y :)$
	Commented by mrW1 last updated on 23/Jul/17

	$the question asks how many times ‘ ‘ 0^{″}$ $is written . i understand for example$ $in the number 200 zero is written 2 times,$ $not one time .$
	Commented by Abbas-Nahi last updated on 23/Jul/17

	$i t s o l v e b y (c o u n t i n g m e t h o d m \times n)$
	Commented by richard last updated on 23/Jul/17

	$w e l l, f o r m e i t m e a n s o n l y t h e n u m b e r s w h e r e t h e z e r o a p p e a r s, i n d e p e n d e n t l y o f t h e a m o u n t o f z e r o s$ $f o r e x e m p l e, i n t h e n u m b e r 1000, e v e n i f t h e r e a r e t h r e e ‘ ‘ 0^{″}, i c o u n t i t o n l y a s o n e n u m b e r w h e r e t h e r e a r e z e r o s, {t h a t}^{'} s m y p o i n t o f v i e w$
	Answered by richard last updated on 22/Jul/17

	$f r o m 1 t o 100 \to 10 z e r o s (10, 20, 30 . . . 100)$ $f r o m 101 t o 200 \to 19 z e r o s (10 z e r o s f r o m 101 t o 110 p l u s 9 z e r o s f r o m 111 t o + 100200)$ $c o n s i d e r i n g t h i s p a d r o n f o r t h e h u n d r e d s t o t h e t h o u s a n d s,$ $\to 101 t o 1000 \equiv 19 \cdot 9 = 181 z e r o s$ $f o r t h e t h o u s a n d s, w e c o u l d s a y t h a t 1001 t o 1100 = 100 z e r o s (1001 t o 1100)$ $A s b e f o r e, f r o m 1100 t o 2000 = 19 \cdot 9 = 181$ $t h i s w a y, k n o w i n g t h i s p a d r o n f o r$ $d o z e n s, h u n d r e d s a n d t h o u s a n d s :$ $2000 \to 2100 = 100 z e r o s$ $2101 \to 3000 = 181 z e r o s$ $3001 \to 3100 = 100 z e r o s$ $3101 \to 3300 = 19 \cdot 2 = 38 z e r o s$ $3301 \to 3333 = 10 + 2 = 12 z e r o s$ $t o t a l :$ $Σ = 10 + 181 + 100 + 181 + 100 + 181 + 100 + 38 + 12$ $Σ = 100 \cdot 3 + 181 \cdot 3 + 10 + 38 + 12$ $Σ = 300 + 543 + 60$ $Σ = 903 z e r o s$
	Answered by Abbas-Nahi last updated on 23/Jul/17

	Answered by mrW1 last updated on 24/Jul/17

	$N = number of numbers in which zero (s) occurs$ $Z = number of zero (s) in the numbers$ $X \in [1, 9]$ $Y \in [1, 3]$ $W \in [1, 2]$ $(1) numbers with 1 digit : X$ $N = 0$ $Z = 0$ $(2) numbers with 2 digits : X 0$ $N = 9$ $Z = 9$ $(3) numbers with 3 digits :$ $(3.1) X 0 X, XX 0$ $N = 2 \times 9 \times 9 = 162$ $Z = 162$ $(3.2) X 00$ $N = 9$ $Z = 9 \times 2 = 18$ $(4) numbers with 4 digits :$ $(4.1) W 0 XX, WX 0 X, WXX 0$ $N = 3 \times 2 \times 9 \times 9 = 486$ $Z = 486$ $(4.2) W 00 X, W 0 X 0, WX 00$ $N = 3 \times 2 \times 9 = 54$ $Z = 54 \times 2 = 108$ $(4.3) W 000$ $N = 2$ $Z = 2 \times 3 = 6$ $(4.4) 30 XX, 3 Y 0 X, 3 WX 0, 33 Y 0$ $N = 9 \times 9 + 3 \times 9 + 2 \times 9 + 3 = 129$ $Z = 129$ $(4.5) 300 X, 3 Y 00, 30 X 0$ $N = 9 + 3 + 9 = 21$ $Z = 21 \times 2 = 42$ $(4.6) 3000$ $N = 1$ $Z = 3$ $Σ N = 9 + 162 + 9 + 486 + 54 + 2 + 129 + 21 + 1 = 873$ $Σ Z = 9 + 162 + 18 + 486 + 108 + 6 + 129 + 42 + 3 = 963$ $i . e . there are 873 numbers in which ‘ ‘ 0^{″} occurs .$ $in them ‘ ‘ 0^{″} is written 963 times .$
	Commented by Tinkutara last updated on 24/Jul/17

	$Thanks Sir! Amazing!$
	Commented by mrW1 last updated on 24/Jul/17

	$Thank you sir for checking!$
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