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If one writes a list of integers from 1 to 1000, then how

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If one writes a list of integers from 1 to 1000, then how [#permalink] New post 08 Jul 2004, 04:35
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If one writes a list of integers from 1 to 1000, then how many zeroes will there be on the list?

Last edited by stolyar on 08 Jul 2004, 05:15, edited 1 time in total.
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 [#permalink] New post 08 Jul 2004, 05:54
191

Consider numbers as
X - 0 (No zeros for this digit)
XX - 9(Total 9 times zero will appear)
XXX - 90(Zero at tens place)+90(Zero at units place) - 1(common case when 0 apears both at tens and units place) = 179
1000 = 3
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 [#permalink] New post 08 Jul 2004, 10:21
147?

0 as unit --> 9 possibilities for tenth & 8 possibilities for huns -->72 possibilities
0 as tenth --> 72 possibilities

72 + 72 + 3(for 1000) = 147
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 [#permalink] New post 08 Jul 2004, 11:25
Two digit Numbers (9)(1) =9

Three digit Numbers (9)(10)(1) + (9)(1)(10) = 180

For 1000 ==> 3

Ans : 192
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 [#permalink] New post 10 Jul 2004, 01:24
ok i might have a different soln

1-100
(10,20,30,40,50,60,70,80,90,100) = 11zeros

1-900 ==> 11 x 9 = 99 zeros

900-1000 ==> 12 zeros
101-109= 9 zeros
201...301...= 9x9 = 81 zeros

99+12+81=192zeros
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 [#permalink] New post 14 Jul 2004, 02:30
Stolyar: What's the OA,

(Also If I am wrong, can anybody point out the mistake.)
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 [#permalink] New post 14 Jul 2004, 07:47
mba wrote:
191

Consider numbers as
X - 0 (No zeros for this digit)
XX - 9(Total 9 times zero will appear)
XXX - 90(Zero at tens place)+90(Zero at units place) - 1(common case when 0 apears both at tens and units place) = 179
1000 = 3


I had made the same mistake in challenge 1 what you have done above.
Since this question is asking how many times 0 is listed we have to account for each 0 that appears.

X00 has 2 zeros. If you remove the common case you will be counting this number only once, but we are not supposed to count the number. We are supposed to count the number zeroes.

X00 as 2 zeros and should be counted twice. 192 should be the correct answer.

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 [#permalink] New post 15 Jul 2004, 01:55
Thanks! ashkg?

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 [#permalink] New post 15 Jul 2004, 11:46
I liked this answer
Two digit Numbers (9)(1) =9

Three digit Numbers (9)(10)(1) + (9)(1)(10) = 180

For 1000 ==> 3

Ans : 192

But, What do
(9)(10)(1) represent

and
(9)(1)(10)
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 [#permalink] New post 16 Jul 2004, 23:44
1- 99 = 9

100 - 199 = 100 -110 + till 190 = 12 + 8

100 - 990 = 20 * 9 = 180

1000 = 3

---------------

192

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 [#permalink] New post 19 Jul 2004, 20:48
I feel that the answer has to be 183 because -

1 - 9 = 0
10 - 99 = 9
100 - 999 = 171 = 9 * 10 *1 + 9 * 1 *10 - 9 ( -9 because we are accounting for 0 in the tenth and unit place twice for nine combinations)
1000 = 3

hence total = 9 + 171 + 3 = 183.

What is the OA stolyar?
  [#permalink] 19 Jul 2004, 20:48
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