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(#M40036055) ELITMUS Logical Reasoning question quant Keep an EYE

number of zeros b/w 1to 999??

Asked In Elitmus amit kumar singh (11 years ago)

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RAKESH  (11 years ago)

ans: 189

no. of zeroes in 2 digit no.s => (_)(0) => 9 no.s => 9 zeroes

in 3 digit no.s

(_)(_)(0) => 9*9=81 n0.s => 81 zeroes
(_)(0)(_) => 9*9=81 no.s => 81 zeroes
(_)(0)(0) => 9 no.s => 9*2=18 zeroes

total number of zeros b/w 1 to 999 = 9 + 81 + 81 + 18 = 189

Is this solution Helpfull? Yes (72) | No (2)  

parul jain  (11 years ago)

Ans 189
1-100=11 0's, 101-200=20 0's similarly
Than 101-900=160
And 901-999=18
Now 11+160+18=189

Is this solution Helpfull? Yes (10) | No   

Mani kandan  (11 years ago)

ans: 189

no. of zeroes in 2 digit no.s => (1)(0),(2)(0),…(9)(0) => 9 no.s => 9 zeroes

in 3 digit no.s

(1)(1)(0),(1)(2)(0),……(9)(9)(0) => 9*9=81 n0.s => 81 zeroes
(1)(0)(1),(1)(0)(2),……(9)(0)(9) => 9*9=81 no.s => 81 zeroes
(1)(0)(0),(2)(0)(0),…….(9)(0)(0) => 9 no.s => 9*2=18 zeroes

total number of zeros b/w 1 to 999 = 9 + 81 + 81 + 18 = 189

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suhaib  (11 years ago)

246
to determine the no of zeroes in a factorial no,,we divide it by 5 and the quotient gives us the no of zeroes
here 1-999=999! so on dividing 999 by 5 we get 246,,actually we only get 995,,we take the nearest value

Is this solution Helpfull? Yes (1) | No (9)  

Gourango  (10 years ago)

ans:189

no of zeros between 1 to 999

at the end digit: (990-10)/10+1= " 99 "

at the middle digit:
extra 1 zeros for 100,200,300.....900 total " 9 " which are not included in 99
and
extra (9 zeros in 101,102.......109)+(9 zeros in 201,202...209)+.....+(9 zeros in 901,902...909)

=9*9= " 81 " zeros

so total= 99 +9 +81= 189

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Arvind Kumar  (10 years ago)

10,20,.....100=11 zeros
lly,110,120......200=11 zeros
...
.
.
.
910,920,.....990=10
so,11*9+10=109

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shaik khader  (11 years ago)

zeros in the two digit number _ _ if u fix 0 at units place then there will be total 9 numbers 9 zeros,similarly for three digit number _ _ _ fix zero at zero at units place first then remaining places can be filled in (9*9)=81 ways so *! zeros illy when zero is at tens place (9*9)=81 numbers 81 zeros ,now _ _ _ zero can be at two positions units and tens so (numbers are possible so total zeros are 9+81+81+18=189 will be the answer

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Roopa Gnanashekar  (11 years ago)

10,20,30,40,50,60,70,80,90,100 so no. of zeroes are 11
101....110 is 10 then 120,130,140,150,160,170,180,190,200 is 10
similarly for 201 to 300 is 20 etc till 801 to 900
901..910 is 10 then 920,930,940,950,960,970,980,990
no. of zeroes =11+8*20+19=11+160+18=189

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Auromika Parida  (11 years ago)

9*1+9*9*1+9*1*9+9*2!=9+81+81+18=189

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kumar pushkal  (10 years ago)

ans:99
n0. of 0's b/w 0 to 99 is 10
so total no. of 0's b/w 0 to 999 is 100
and question asks 1 to 999 so 100-1= 99

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rahul saini  (8 years ago)

from 1-100 (10,20,30,40,50,60,70,80,90,100) no. of zeroes are 11
from 101-200 (101-110 0's --->10 + 120..200 0's ----> 10) no. of zeroes are 20
from 101-200 (201-210 0's --->10 + 220..300 0's ----> 10) no. of zeroes are 20
from 101-200 (301-310 0's --->10 + 320..400 0's ----> 10) no. of zeroes are 20
from 101-200 (401-410 0's --->10 + 420..500 0's ----> 10) no. of zeroes are 20
from 101-200 (501-510 0's --->10 + 520..600 0's ----> 10) no. of zeroes are 20
from 101-200 (601-610 0's --->10 + 620..700 0's ----> 10) no. of zeroes are 20
from 101-200 (701-710 0's --->10 + 720..800 0's ----> 10) no. of zeroes are 20
from 101-200 (801-810 0's --->10 + 820..900 0's ----> 10) no. of zeroes are 20
from 101-200 (901-910 0's --->10 + 920..990 0's ----> 08) no. of zeroes are 18
so total zeroes are 11+(8*20)+18 = 189

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Aqib Azizkhan  (8 years ago)

1 to 100=11
101 to 200=20
201 to 300=20
301 to 400=20
401 to 500=20
501 to 600=20
601 to 700=20
701 to 800=20
801 to 900=20
901 to 999=18
total zeros=11+(20*8)+18=189

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