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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?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{times}\:\mathrm{is}\:\mathrm{digit}\:\mathrm{0}\:\mathrm{written}\:\mathrm{when} \\ $$$$\mathrm{listing}\:\mathrm{all}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:\mathrm{3333}? \\ $$
Commented by richard last updated on 23/Jul/17
do you want how many zeros there are between 1 and 3333, or how many times the “0” appears, independently of the amount of zeros?
$${do}\:{you}\:{want}\:{how}\:{many}\:{zeros}\:{there}\:{are}\:{between}\:\mathrm{1}\:{and}\:\mathrm{3333},\:{or}\:{how}\:{many}\:{times}\:{the}\:“\mathrm{0}''\:{appears},\:{independently}\:{of}\:{the}\:{amount}\:{of}\:{zeros}? \\ $$
Answered by Abbas-Nahi last updated on 23/Jul/17
I thin the times of digit is 9×10×10×1=900
$$\mathrm{I}\:{thin}\:{the}\:{times}\:{of}\:{digit}\:{is} \\ $$$$\mathrm{9}×\mathrm{10}×\mathrm{10}×\mathrm{1}=\mathrm{900} \\ $$
Answered by richard last updated on 23/Jul/17
from 1 to 100→10 times(10,20,30...100) from 101 to 200→19 times(10 times the zero(or zeros appears) from 101 to 110 plus 9 times the zero appears from 111 to 1000) considering this padron for the hundreds to the thousands, →101 to 1000≡ 19∙9=181 times for the thousands, we could say that for 1001 to 1100= 100 times (the zero or more than one zero appears) As before, from 1100 to 2000=19∙9=181 this way, knowing this padronfor dozens, hundreds and thousands: 2000→2100=100 times 2101→3000= 181 times 3001→3100=100 times 3101→3300=19∙2= 38 times 3301→3333=10+2=12 times total: Σ=10+181+100+181+100+181+100+38+12 Σ=100∙3+181∙3+10+38+12 Σ=300+543+60 Σ=903 times ( once more, when i say “times”, i mean the number of times the zero or more than one zero appears between 1 and 3333)
$${from}\:\mathrm{1}\:{to}\:\mathrm{100}\rightarrow\mathrm{10}\:{times}\left(\mathrm{10},\mathrm{20},\mathrm{30}…\mathrm{100}\right) \\ $$$${from}\:\mathrm{101}\:{to}\:\mathrm{200}\rightarrow\mathrm{19}\:{times}\left(\mathrm{10}\:{times}\:{the}\:{zero}\left({or}\:{zeros}\:{appears}\right)\:{from}\:\mathrm{101}\:{to}\:\mathrm{110}\:{plus}\:\mathrm{9}\:{times}\:{the}\:{zero}\:{appears}\:{from}\:\mathrm{111}\:{to}\:\mathrm{1000}\right) \\ $$$${considering}\:{this}\:{padron}\:{for}\:{the}\:{hundreds}\:{to}\:{the}\:{thousands}, \\ $$$$\rightarrow\mathrm{101}\:{to}\:\mathrm{1000}\equiv\:\mathrm{19}\centerdot\mathrm{9}=\mathrm{181}\:{times} \\ $$$${for}\:{the}\:{thousands},\:{we}\:{could}\:{say}\:{that}\:{for}\:\mathrm{1001}\:{to}\:\mathrm{1100}=\:\mathrm{100}\:{times}\:\left({the}\:{zero}\:{or}\:{more}\:{than}\:{one}\:{zero}\:{appears}\right) \\ $$$${As}\:{before},\:{from}\:\mathrm{1100}\:{to}\:\mathrm{2000}=\mathrm{19}\centerdot\mathrm{9}=\mathrm{181} \\ $$$${this}\:{way},\:{knowing}\:{this}\:{padronfor} \\ $$$${dozens},\:{hundreds}\:{and}\:{thousands}: \\ $$$$\mathrm{2000}\rightarrow\mathrm{2100}=\mathrm{100}\:{times} \\ $$$$\mathrm{2101}\rightarrow\mathrm{3000}=\:\mathrm{181}\:{times} \\ $$$$\mathrm{3001}\rightarrow\mathrm{3100}=\mathrm{100}\:{times} \\ $$$$\mathrm{3101}\rightarrow\mathrm{3300}=\mathrm{19}\centerdot\mathrm{2}=\:\mathrm{38}\:{times} \\ $$$$\mathrm{3301}\rightarrow\mathrm{3333}=\mathrm{10}+\mathrm{2}=\mathrm{12}\:{times} \\ $$$${total}: \\ $$$$\Sigma=\mathrm{10}+\mathrm{181}+\mathrm{100}+\mathrm{181}+\mathrm{100}+\mathrm{181}+\mathrm{100}+\mathrm{38}+\mathrm{12} \\ $$$$\Sigma=\mathrm{100}\centerdot\mathrm{3}+\mathrm{181}\centerdot\mathrm{3}+\mathrm{10}+\mathrm{38}+\mathrm{12} \\ $$$$\Sigma=\mathrm{300}+\mathrm{543}+\mathrm{60} \\ $$$$\Sigma=\mathrm{903}\:{times}\:\left(\:{once}\:{more},\:{when}\:{i}\:{say}\:“{times}'',\:{i}\:{mean}\:{the}\:{number}\:{of}\:{times}\:{the}\:{zero}\:{or}\:{more}\:{than}\:{one}\:{zero}\:{appears}\:{between}\:\mathrm{1}\:{and}\:\mathrm{3333}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mrW1 last updated on 23/Jul/17
from 1 to 100→11 zeros(10,20,30...100)
$${from}\:\mathrm{1}\:{to}\:\mathrm{100}\rightarrow\mathrm{11}\:{zeros}\left(\mathrm{10},\mathrm{20},\mathrm{30}…\mathrm{100}\right) \\ $$$$ \\ $$
Commented by richard last updated on 23/Jul/17
why 11? the question asks about how many times the zero appears, not how many zeros exists between then, is it not?
$${why}\:\mathrm{11}?\:{the}\:{question}\:{asks}\:{about}\:{how}\:{many}\:{times}\:{the}\:{zero}\:{appears},\:{not}\:{how}\:{many}\:{zeros}\:{exists}\:{between}\:{then},\:{is}\:{it}\:{not}? \\ $$
Commented by richard last updated on 23/Jul/17
I need to fix it, thd correct answer is Σ=903 times It means the zero(or more zeros)r appears 903 times between 1 and 3333, not how many zeros are between 1 and 3333
$${I}\:{need}\:{to}\:{fix}\:{it},\:{thd}\:{correct}\:{answer}\:{is} \\ $$$$\Sigma=\mathrm{903}\:{times} \\ $$$${It}\:{means}\:{the}\:{zero}\left({or}\:{more}\:{zeros}\right){r}\:{appears}\:\mathrm{903}\:{times}\:{between}\:\mathrm{1}\:{and}\:\mathrm{3333},\:{not}\:{how}\:{many}\:{zeros}\:{are}\:{between}\:\mathrm{1}\:{and}\:\mathrm{3333} \\ $$
Commented by richard last updated on 23/Jul/17
the∗
$${the}\ast \\ $$
Commented by richard last updated on 23/Jul/17
It should be Σ=903 times, not Σ=903 zeros, sorry :)
$$\left.{It}\:{should}\:{be}\:\Sigma=\mathrm{903}\:{times},\:{not}\:\Sigma=\mathrm{903}\:{zeros},\:{sorry}\::\right) \\ $$
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.
$$\mathrm{the}\:\mathrm{question}\:\mathrm{asks}\:\mathrm{how}\:\mathrm{many}\:\mathrm{times}\:“\mathrm{0}'' \\ $$$$\mathrm{is}\:\mathrm{written}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{for}\:\mathrm{example} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{number}\:\mathrm{200}\:\mathrm{zero}\:\mathrm{is}\:\mathrm{written}\:\mathrm{2}\:\mathrm{times}, \\ $$$$\mathrm{not}\:\mathrm{one}\:\mathrm{time}. \\ $$
Commented by Abbas-Nahi last updated on 23/Jul/17
it solve by (counting method m×n)
$${it}\:{solve}\:{by}\:\left({counting}\:{method}\:\:{m}×{n}\right) \\ $$
Commented by richard last updated on 23/Jul/17
well, for me it means only the numbers where the zero appears, independently of the amountof zeros for exemple, inthe number 1000, even if there are three “0”, i count it only as one number where there are zeros, that′s my point of view
$$ \\ $$$${well},\:{for}\:{me}\:{it}\:{means}\:{only}\:{the}\:{numbers}\:{where}\:{the}\:{zero}\:{appears},\:{independently}\:{of}\:{the}\:{amountof}\:{zeros} \\ $$$${for}\:{exemple},\:{inthe}\:{number}\:\mathrm{1000},\:{even}\:{if}\:{there}\:{are}\:{three}\:“\mathrm{0}'',\:{i}\:{count}\:{it}\:{only}\:{as}\:{one}\:{number}\:{where}\:{there}\:{are}\:{zeros},\:{that}'{s}\:{my}\:{point}\:{of}\:{view} \\ $$$$ \\ $$
Answered by richard last updated on 22/Jul/17
from 1 to 100→10 zeros(10,20,30...100) from 101 to 200→19 zeros(10 zeros from 101 to 110 plus 9 zeros from 111 to +100200) considering this padron for the hundreds to the thousands, →101 to 1000≡ 19∙9=181 zeros for the thousands, we could say that 1001 to 1100= 100 zeros(1001 to 1100) As before, from 1100 to 2000=19∙9=181 this way, knowing this padronfor dozens, hundreds and thousands: 2000→2100=100 zeros 2101→3000= 181 zeros 3001→3100=100 zeros 3101→3300=19∙2= 38 zeros 3301→3333=10+2=12 zeros total: Σ=10+181+100+181+100+181+100+38+12 Σ=100∙3+181∙3+10+38+12 Σ=300+543+60 Σ=903 zeros
$${from}\:\mathrm{1}\:{to}\:\mathrm{100}\rightarrow\mathrm{10}\:{zeros}\left(\mathrm{10},\mathrm{20},\mathrm{30}…\mathrm{100}\right) \\ $$$${from}\:\mathrm{101}\:{to}\:\mathrm{200}\rightarrow\mathrm{19}\:{zeros}\left(\mathrm{10}\:{zeros}\:{from}\:\mathrm{101}\:{to}\:\mathrm{110}\:{plus}\:\mathrm{9}\:{zeros}\:{from}\:\mathrm{111}\:{to}\:+\mathrm{100200}\right) \\ $$$${considering}\:{this}\:{padron}\:{for}\:{the}\:{hundreds}\:{to}\:{the}\:{thousands}, \\ $$$$\rightarrow\mathrm{101}\:{to}\:\mathrm{1000}\equiv\:\mathrm{19}\centerdot\mathrm{9}=\mathrm{181}\:{zeros} \\ $$$${for}\:{the}\:{thousands},\:{we}\:{could}\:{say}\:{that}\:\mathrm{1001}\:{to}\:\mathrm{1100}=\:\mathrm{100}\:{zeros}\left(\mathrm{1001}\:{to}\:\mathrm{1100}\right) \\ $$$${As}\:{before},\:{from}\:\mathrm{1100}\:{to}\:\mathrm{2000}=\mathrm{19}\centerdot\mathrm{9}=\mathrm{181} \\ $$$${this}\:{way},\:{knowing}\:{this}\:{padronfor} \\ $$$${dozens},\:{hundreds}\:{and}\:{thousands}: \\ $$$$\mathrm{2000}\rightarrow\mathrm{2100}=\mathrm{100}\:{zeros} \\ $$$$\mathrm{2101}\rightarrow\mathrm{3000}=\:\mathrm{181}\:{zeros} \\ $$$$\mathrm{3001}\rightarrow\mathrm{3100}=\mathrm{100}\:{zeros} \\ $$$$\mathrm{3101}\rightarrow\mathrm{3300}=\mathrm{19}\centerdot\mathrm{2}=\:\mathrm{38}\:{zeros} \\ $$$$\mathrm{3301}\rightarrow\mathrm{3333}=\mathrm{10}+\mathrm{2}=\mathrm{12}\:{zeros} \\ $$$${total}: \\ $$$$\Sigma=\mathrm{10}+\mathrm{181}+\mathrm{100}+\mathrm{181}+\mathrm{100}+\mathrm{181}+\mathrm{100}+\mathrm{38}+\mathrm{12} \\ $$$$\Sigma=\mathrm{100}\centerdot\mathrm{3}+\mathrm{181}\centerdot\mathrm{3}+\mathrm{10}+\mathrm{38}+\mathrm{12} \\ $$$$\Sigma=\mathrm{300}+\mathrm{543}+\mathrm{60} \\ $$$$\Sigma=\mathrm{903}\:{zeros} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
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∈[1,9] Y∈[1,3] W∈[1,2] (1) numbers with 1 digit: X N=0 Z=0 (2) numbers with 2 digits: X0 N=9 Z=9 (3) numbers with 3 digits: (3.1) X0X, XX0 N=2×9×9=162 Z=162 (3.2) X00 N=9 Z=9×2=18 (4) numbers with 4 digits: (4.1) W0XX, WX0X, WXX0 N=3×2×9×9=486 Z=486 (4.2) W00X, W0X0, WX00 N=3×2×9=54 Z=54×2=108 (4.3) W000 N=2 Z=2×3=6 (4.4) 30XX, 3Y0X, 3WX0, 33Y0 N=9×9+3×9+2×9+3=129 Z=129 (4.5) 300X, 3Y00, 30X0 N=9+3+9=21 Z=21×2=42 (4.6) 3000 N=1 Z=3 ΣN=9+162+9+486+54+2+129+21+1=873 ΣZ=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.
$$\mathrm{N}=\mathrm{number}\:\mathrm{of}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{which}\:\mathrm{zero}\left(\mathrm{s}\right)\:\mathrm{occurs} \\ $$$$\mathrm{Z}=\mathrm{number}\:\mathrm{of}\:\mathrm{zero}\left(\mathrm{s}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{numbers} \\ $$$$\mathrm{X}\in\left[\mathrm{1},\mathrm{9}\right] \\ $$$$\mathrm{Y}\in\left[\mathrm{1},\mathrm{3}\right] \\ $$$$\mathrm{W}\in\left[\mathrm{1},\mathrm{2}\right] \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{1}\:\mathrm{digit}:\:\mathrm{X} \\ $$$$\mathrm{N}=\mathrm{0} \\ $$$$\mathrm{Z}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{2}\:\mathrm{digits}:\:\mathrm{X0} \\ $$$$\mathrm{N}=\mathrm{9} \\ $$$$\mathrm{Z}=\mathrm{9} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{3}\:\mathrm{digits}:\: \\ $$$$\left(\mathrm{3}.\mathrm{1}\right)\:\mathrm{X0X},\:\mathrm{XX0} \\ $$$$\mathrm{N}=\mathrm{2}×\mathrm{9}×\mathrm{9}=\mathrm{162} \\ $$$$\mathrm{Z}=\mathrm{162} \\ $$$$\left(\mathrm{3}.\mathrm{2}\right)\:\mathrm{X00} \\ $$$$\mathrm{N}=\mathrm{9} \\ $$$$\mathrm{Z}=\mathrm{9}×\mathrm{2}=\mathrm{18} \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{4}\:\mathrm{digits}:\: \\ $$$$\left(\mathrm{4}.\mathrm{1}\right)\:\mathrm{W0XX},\:\mathrm{WX0X},\:\mathrm{WXX0} \\ $$$$\mathrm{N}=\mathrm{3}×\mathrm{2}×\mathrm{9}×\mathrm{9}=\mathrm{486} \\ $$$$\mathrm{Z}=\mathrm{486} \\ $$$$\left(\mathrm{4}.\mathrm{2}\right)\:\mathrm{W00X},\:\mathrm{W0X0},\:\mathrm{WX00} \\ $$$$\mathrm{N}=\mathrm{3}×\mathrm{2}×\mathrm{9}=\mathrm{54} \\ $$$$\mathrm{Z}=\mathrm{54}×\mathrm{2}=\mathrm{108} \\ $$$$\left(\mathrm{4}.\mathrm{3}\right)\:\mathrm{W000} \\ $$$$\mathrm{N}=\mathrm{2} \\ $$$$\mathrm{Z}=\mathrm{2}×\mathrm{3}=\mathrm{6} \\ $$$$\left(\mathrm{4}.\mathrm{4}\right)\:\mathrm{30XX},\:\mathrm{3Y0X},\:\mathrm{3WX0},\:\mathrm{33Y0} \\ $$$$\mathrm{N}=\mathrm{9}×\mathrm{9}+\mathrm{3}×\mathrm{9}+\mathrm{2}×\mathrm{9}+\mathrm{3}=\mathrm{129} \\ $$$$\mathrm{Z}=\mathrm{129} \\ $$$$\left(\mathrm{4}.\mathrm{5}\right)\:\mathrm{300X},\:\mathrm{3Y00},\:\mathrm{30X0} \\ $$$$\mathrm{N}=\mathrm{9}+\mathrm{3}+\mathrm{9}=\mathrm{21} \\ $$$$\mathrm{Z}=\mathrm{21}×\mathrm{2}=\mathrm{42} \\ $$$$\left(\mathrm{4}.\mathrm{6}\right)\:\mathrm{3000} \\ $$$$\mathrm{N}=\mathrm{1} \\ $$$$\mathrm{Z}=\mathrm{3} \\ $$$$ \\ $$$$\Sigma\mathrm{N}=\mathrm{9}+\mathrm{162}+\mathrm{9}+\mathrm{486}+\mathrm{54}+\mathrm{2}+\mathrm{129}+\mathrm{21}+\mathrm{1}=\mathrm{873} \\ $$$$\Sigma\mathrm{Z}=\mathrm{9}+\mathrm{162}+\mathrm{18}+\mathrm{486}+\mathrm{108}+\mathrm{6}+\mathrm{129}+\mathrm{42}+\mathrm{3}=\mathrm{963} \\ $$$$ \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{there}\:\mathrm{are}\:\mathrm{873}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{which}\:“\mathrm{0}''\:\mathrm{occurs}. \\ $$$$\mathrm{in}\:\mathrm{them}\:“\mathrm{0}''\:\mathrm{is}\:\mathrm{written}\:\mathrm{963}\:\mathrm{times}. \\ $$
Commented by Tinkutara last updated on 24/Jul/17
Thanks Sir! Amazing!
$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{Amazing}! \\ $$
Commented by mrW1 last updated on 24/Jul/17
Thank you sir for checking!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{checking}! \\ $$

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