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Re: Counting digits (m09q17) [#permalink]
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01 Sep 2010, 03:05
what would be the difficulty level for this question in real GMAT ?
i have a request. i dont know if many of us feel the same way. while posting the question is it possible to mention the difficulty level in the real GMAT also ? that would be really helpful...



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Re: Counting digits (m09q17) [#permalink]
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01 Sep 2010, 07:10
Not many people really know the difficulty level since the authors of the question don't really know. One way we can tell the difficulty is if the question comes from the OG because the easy questions are in the front of a section and become more difficult as the question numbers get higher. This particular question is not very difficult. Maybe around the 575  625 range? Anyone else want to give their opinion as to the difficulty level of this question? srivicool wrote: what would be the difficulty level for this question in real GMAT ?
i have a request. i dont know if many of us feel the same way. while posting the question is it possible to mention the difficulty level in the real GMAT also ? that would be really helpful...
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Re: Counting digits (m09q17) [#permalink]
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01 Sep 2010, 09:38
thanks jallenmorris for the response. i got your point..



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Re: Counting digits (m09q17) [#permalink]
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02 Sep 2010, 03:49
Facts except 0 every digit from 19 comes about 19 times (need to remember only once) further apply simple mathematical formula......19*7=133



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Re: Counting digits [#permalink]
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02 Sep 2010, 16:24
Oski wrote: Actually you don't have to make "slices" (19, 1099, 100  999)
For instance in your first question "How many times will the digit 7 written when listing integers from 1 to 1000 ?" :
 Count the number of times 7 appears as a digits number : once every ten, that's 1000/10 = 100 times
 Count the number of times 7 appears as a tens number : ten times every hundred, that's 1000/100*10 = 100 times
 Count the number of times 7 appears as a hundreds number : a hundred times every thousand, that's 1000/1000*100 = 100 times
==> Total = 100 + 100 + 100 = 300 times
For your second question "on a book of 705 pages on how many pages does digit 9 appear in the numeration?", and as it was said above, you have to be careful not to double count.
First, notice it is the same than asking "on a book of 700 pages on how many pages does digit 9 appear in the numeration?" (since 701, 702, 703, 704 and 705 don't contain 9s)
Then same method as above :
 Count the number of times 9 appears as a digits number: once every ten, that's 700/10 = 70 times
 Count the number of times 9 appears as a tens number and not as a digits number: 9 times every hundred, that's 700/100*9 = 63 times
 Count the number of times 9 appears as a hundreds number: zero since there are less than 900 pages
==> Total = 70 + 63 + 0 = 133 pages Very nice explanation for the first question. How would you, however, count the numbers containing 7, not how many times will 7 appear in this 11000 set?
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Re: Counting digits (m09q17) [#permalink]
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17 Sep 2010, 08:14
But 300 isn't even an option in the original multiple choice answers. Pkit wrote: jallenmorris wrote: Not many people really know the difficulty level since the authors of the question don't really know. One way we can tell the difficulty is if the question comes from the OG because the easy questions are in the front of a section and become more difficult as the question numbers get higher.
This particular question is not very difficult. Maybe around the 575  625 range? Anyone else want to give their opinion as to the difficulty level of this question?
I think it is of 600650 level. Here is a smart shortcut. in range from 10 to 99  you have 11 figures in "own line" and 8 "out of own line" = total 19 (7th's own line is 70 to 79) in range from 100 to 1000  you have 120 figures in "own line" and 20 "out of own line" = total 280. +1 figure from 1 to 9 total 300.
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Re: Counting digits (m09q17) [#permalink]
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17 Sep 2010, 09:09
I am former big4auditor, 300 is just a reconciliation that my counting is right.
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Re: Counting digits (m09q17) [#permalink]
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17 Sep 2010, 09:12
Sorry, I'm not an auditory. I'm an attorney so I only think in terms of Guilty/Innocent and Liabilities. Pkit wrote: :) I am former big4auditor, 300 is just a reconciliation that my counting is right.
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Re: Counting digits (m09q17) [#permalink]
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11 Aug 2011, 16:23
ok, I was trying for a different method, after rectifying my silly mistake I see that, this problem can be approached in a different way
Here it is
705  ( all cases when 9 doesnt come )
705  [ ( one digit where 9 doesnt come ) + ( two digit where 9 doesnt come ) + (three digit where 9 doesnt come ) ]
705  [ 8 + ( 8 X 9 ) + ( 6 X 9 X 9 ) + 6 ] ( \\ 700705 > 6 )
705  [ 8 + 72 + 486 + 6 ]
= 133



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Re: Counting digits (m09q17) [#permalink]
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11 Aug 2011, 23:37
The problem was easy to solve but real good discussion happening here.. jallenmorris wrote: Sorry, I'm not an auditory. I'm an attorney so I only think in terms of Guilty/Innocent and Liabilities. Pkit wrote: :) I am former big4auditor, 300 is just a reconciliation that my counting is right.
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Re: Counting digits [#permalink]
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05 Sep 2011, 09:10
Oski wrote: Actually you don't have to make "slices" (19, 1099, 100  999)
For instance in your first question "How many times will the digit 7 written when listing integers from 1 to 1000 ?" :
 Count the number of times 7 appears as a digits number : once every ten, that's 1000/10 = 100 times
 Count the number of times 7 appears as a tens number : ten times every hundred, that's 1000/100*10 = 100 times
 Count the number of times 7 appears as a hundreds number : a hundred times every thousand, that's 1000/1000*100 = 100 times
==> Total = 100 + 100 + 100 = 300 times
For your second question "on a book of 705 pages on how many pages does digit 9 appear in the numeration?", and as it was said above, you have to be careful not to double count.
First, notice it is the same than asking "on a book of 700 pages on how many pages does digit 9 appear in the numeration?" (since 701, 702, 703, 704 and 705 don't contain 9s)
Then same method as above :
 Count the number of times 9 appears as a digits number: once every ten, that's 700/10 = 70 times
 Count the number of times 9 appears as a tens number and not as a digits number: 9 times every hundred, that's 700/100*9 = 63 times
 Count the number of times 9 appears as a hundreds number: zero since there are less than 900 pages
==> Total = 70 + 63 + 0 = 133 pages nice way to count...will it always wrk out to be correct??...looks like lot of clever thinking is required in parallel to avoid mistake!!
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Re: Counting digits (m09q17) [#permalink]
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05 Sep 2011, 10:16
Answer D 10 times for each set of 100's (for the 90's) [90, 91, 92....]: 10 x 7 = 70 9 times for each set of 100's for the single occurrences, excluding the 90's, [9, 19, 29...]: 9 x 7 = 63
70+63 = 133



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Re: Counting digits [#permalink]
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05 Sep 2011, 21:44
Its D, Liked this explanation. Oski wrote: Actually you don't have to make "slices" (19, 1099, 100  999)
For instance in your first question "How many times will the digit 7 written when listing integers from 1 to 1000 ?" :
 Count the number of times 7 appears as a digits number : once every ten, that's 1000/10 = 100 times
 Count the number of times 7 appears as a tens number : ten times every hundred, that's 1000/100*10 = 100 times
 Count the number of times 7 appears as a hundreds number : a hundred times every thousand, that's 1000/1000*100 = 100 times
==> Total = 100 + 100 + 100 = 300 times
For your second question "on a book of 705 pages on how many pages does digit 9 appear in the numeration?", and as it was said above, you have to be careful not to double count.
First, notice it is the same than asking "on a book of 700 pages on how many pages does digit 9 appear in the numeration?" (since 701, 702, 703, 704 and 705 don't contain 9s)
Then same method as above :
 Count the number of times 9 appears as a digits number: once every ten, that's 700/10 = 70 times
 Count the number of times 9 appears as a tens number and not as a digits number: 9 times every hundred, that's 700/100*9 = 63 times
 Count the number of times 9 appears as a hundreds number: zero since there are less than 900 pages
==> Total = 70 + 63 + 0 = 133 pages
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Re: Counting digits (m09q17) [#permalink]
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05 Sep 2011, 22:17
Hi All
Lets try a simple math formula on this 1 to 89 no 9's are 9 then from 90 to 99 no of 9's are 10 so 19 so these number will be repeated 7 times 1100, 101200, 201300, 301400, 401500, 501600, 601705
So total pages = 19* 7 = 133



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Re: Counting digits (m09q17) [#permalink]
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06 Sep 2011, 05:33
Did not look at all the replies but I would like to propose a simple solution:
Number of total 9s = Number of 9s at unit place + Number of 9s at tens place = 7*(that b/w 1100) + 7*(that b/w 1100) = 7*(that b/w 90100) +7*10*(that b/w 110) = 7*10 + 7*10*1 = 140
So there are 2 mutually exclusive overlapping sets (pages with 9 at tens place AND pages with 9 at unit place  NOTE: they overlap). Imagine a venn diagram or double matrix  whichever makes it easy for you to visualize.
Number of pages where 1 or 2 9 appears = Number of pages with 9 at tens place + Number of pages with 9 at unit place  Number of pages with 9 at both tens and unit place = 140  7*Number of pages with 9 at tens and unit place b/w 1100 = 140  7*1 = 133
Hope this helps.
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Re: Counting digits (m09q17) [#permalink]
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09 Sep 2011, 13:17
7 X Y  X Y 7  X 7 Y number of times10 109 109 10 1009090 = 280
Total number of times = 280+19+1=300
Why not subtract the repeated numbers such as 777,707,770 in this situation?



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Re: Counting digits (m09q17) [#permalink]
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09 Sep 2011, 14:24
aaron22197 wrote: All pages of the book are numbered. If the first page is numbered 1 and the last page is numbered 705, on how many pages does digit 9 appear in the numeration?
(A) 70 (B) 77 (C) 126 (D) 133 (E) 140 19 9(Total=1) 1099 Total(Numbers without 9)=9*108*9=9072=18 Total(1100)=18+1=19 From, (1700)=19*7=133 Ans: "D"
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Re: Counting digits (m09q17) [#permalink]
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20 Nov 2011, 21:46
19 times every hundred so 19*7 =133 ie D
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Re: Counting digits (m09q17) [#permalink]
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14 Oct 2012, 08:50
Here is another way consider numbers from 1 to 699 since after 699 no number contains 9 now, how many numbers can be formed without using 9 = 7 x 9 x 9 = 567
therefore 700 567= 133




Re: Counting digits (m09q17)
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