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




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?
Howmanytimesisdigit0writtenwhenlistingallnumbersfrom1to3333?
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?
doyouwanthowmanyzerostherearebetween1and3333,orhowmanytimesthe0appears,independentlyoftheamountofzeros?
Answered by Abbas-Nahi last updated on 23/Jul/17
I thin the times of digit is  9×10×10×1=900
Ithinthetimesofdigitis9×10×10×1=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)
from1to10010times(10,20,30100)from101to20019times(10timesthezero(orzerosappears)from101to110plus9timesthezeroappearsfrom111to1000)consideringthispadronforthehundredstothethousands,101to1000199=181timesforthethousands,wecouldsaythatfor1001to1100=100times(thezeroormorethanonezeroappears)Asbefore,from1100to2000=199=181thisway,knowingthispadronfordozens,hundredsandthousands:20002100=100times21013000=181times30013100=100times31013300=192=38times33013333=10+2=12timestotal:Σ=10+181+100+181+100+181+100+38+12Σ=1003+1813+10+38+12Σ=300+543+60Σ=903times(oncemore,whenisaytimes,imeanthenumberoftimesthezeroormorethanonezeroappearsbetween1and3333)
Commented by mrW1 last updated on 23/Jul/17
from 1 to 100→11 zeros(10,20,30...100)
from1to10011zeros(10,20,30100)
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?
why11?thequestionasksabouthowmanytimesthezeroappears,nothowmanyzerosexistsbetweenthen,isitnot?
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
Ineedtofixit,thdcorrectanswerisΣ=903timesItmeansthezero(ormorezeros)rappears903timesbetween1and3333,nothowmanyzerosarebetween1and3333
Commented by richard last updated on 23/Jul/17
the∗
the
Commented by richard last updated on 23/Jul/17
It should be Σ=903 times, not Σ=903 zeros, sorry :)
ItshouldbeΣ=903times,notΣ=903zeros,sorry:)
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.
thequestionaskshowmanytimes0iswritten.iunderstandforexampleinthenumber200zeroiswritten2times,notonetime.
Commented by Abbas-Nahi last updated on 23/Jul/17
it solve by (counting method  m×n)
itsolveby(countingmethodm×n)
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,formeitmeansonlythenumberswherethezeroappears,independentlyoftheamountofzerosforexemple,inthenumber1000,eveniftherearethree0,icountitonlyasonenumberwheretherearezeros,thatsmypointofview
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
from1to10010zeros(10,20,30100)from101to20019zeros(10zerosfrom101to110plus9zerosfrom111to+100200)consideringthispadronforthehundredstothethousands,101to1000199=181zerosforthethousands,wecouldsaythat1001to1100=100zeros(1001to1100)Asbefore,from1100to2000=199=181thisway,knowingthispadronfordozens,hundredsandthousands:20002100=100zeros21013000=181zeros30013100=100zeros31013300=192=38zeros33013333=10+2=12zerostotal:Σ=10+181+100+181+100+181+100+38+12Σ=1003+1813+10+38+12Σ=300+543+60Σ=903zeros
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.
N=numberofnumbersinwhichzero(s)occursZ=numberofzero(s)inthenumbersX[1,9]Y[1,3]W[1,2](1)numberswith1digit:XN=0Z=0(2)numberswith2digits:X0N=9Z=9(3)numberswith3digits:(3.1)X0X,XX0N=2×9×9=162Z=162(3.2)X00N=9Z=9×2=18(4)numberswith4digits:(4.1)W0XX,WX0X,WXX0N=3×2×9×9=486Z=486(4.2)W00X,W0X0,WX00N=3×2×9=54Z=54×2=108(4.3)W000N=2Z=2×3=6(4.4)30XX,3Y0X,3WX0,33Y0N=9×9+3×9+2×9+3=129Z=129(4.5)300X,3Y00,30X0N=9+3+9=21Z=21×2=42(4.6)3000N=1Z=3ΣN=9+162+9+486+54+2+129+21+1=873ΣZ=9+162+18+486+108+6+129+42+3=963i.e.thereare873numbersinwhich0occurs.inthem0iswritten963times.
Commented by Tinkutara last updated on 24/Jul/17
Thanks Sir! Amazing!
ThanksSir!Amazing!
Commented by mrW1 last updated on 24/Jul/17
Thank you sir for checking!
Thankyousirforchecking!

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