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# Counting digits (m09q17)

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07 Jul 2008, 18:31
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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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

I have always had trouble with "counting digits" problems
I struggled with this problem during a test(it took me 3.5 mins to crack it)

How many times will the digit 7 written when listing integers from 1 to 1000 ?

My standard approach is- I try to figure out in how many ways can each slot be occupied

1 time between 1 and 10
19 times between 1 and 100

For 3 digit numbers

----------------------7 X Y ---|-- X Y 7 --|-- X 7 Y
number of times----10 10------9 10---------9 10
-------------------------100---------90------------90 = 280

Total number of times = 280+19+1=300

So , I was happy with this above strategy and I thought I could apply it to any counting problem.

Today, I encountered a fairly simple question (up).

I used the simple counting method to to 19*7 = 133
However, I could not get the same answer by following the X Y Z strategy shown above.

Two questions for the experts

A)What am I doing wrong in the 2nd problem ?
B) Is there a better approach than what I am doing?

Any help appreciated...
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08 Jul 2008, 09:12
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Actually you don't have to make "slices" (1-9, 10-99, 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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07 Jul 2008, 19:51
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I'll bet you got 140 when you used the XYZ method above. Here's why.

The method you use above will count the number of 9's from 1 to 705.

9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98,99..ah ha!!! That counts 2, but there is only 1 page, as the question asks.

When I wrote out the numbers (in excel, not by hand ), I got 133, but was also confused because I see the value to your method.

The key is that when you're counting the number of 9's in the digits and the number of 9's in the ten's column, you must subtract out the 99, 199, 299, 399, 499, 599, 699 because that's only 1 page, but two 9's on the page.

So the units & tens method is:

10-9's per 100 numbres x 7 (up to 705 by 100 is 7 units of 100)...so 70 for the units.

the Tens column is all of the 90s, then 190's,etc so 10 per 100 there too. Up to 705, that's 7 units of 100 also, so 70. I see a total of 140-9's. But we must now subtract out the numbers with duplicate 9's on it.

99, 199, 299, 399, 499, 599, 699...there are 7 of them. 140 - 7 = 133....a.k.a, the answer!
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 22 May 2007 Posts: 217 Followers: 2 Kudos [?]: 82 [1] , given: 0 Re: Counting digits [#permalink] ### Show Tags 07 Jul 2008, 20:06 1 This post received KUDOS Thanks Jallen +1. That makes sense. It is not counting the number of 9's but counting the number of pages that contain a 9. I'm glad the XYZ method is still valid for other counting type problems. I should have read the question better. Manager Joined: 04 Aug 2010 Posts: 158 Followers: 2 Kudos [?]: 27 [1] , given: 15 Re: Counting digits (m09q17) [#permalink] ### Show Tags 31 Aug 2010, 13:42 1 This post received KUDOS I got D. (9,19,29,39,49,59,69,79,89,90,91,92,93,94,5,96,97,98,99) X 6 (for 0xx,1xx,2xx,3xx,4xx,5xx,6xx) Manager Joined: 27 Jul 2010 Posts: 197 Location: Prague Schools: University of Economics Prague Followers: 1 Kudos [?]: 41 [1] , given: 15 Re: Counting digits (m09q17) [#permalink] ### Show Tags 02 Sep 2010, 15:10 1 This post received KUDOS 19*7 = 133 that's clear. The problem for me was how easily come to 19. My method is maybe stupid, but it is quick and works: 1-10 - one*9 11-20 - two*9 The number of nines is exaclty the number of decimal digit + 1. Therefore 90 there are 9+1 nines = 10. now only add 9 nines from the 91-99 and Voilà: we have 19. _________________ You want somethin', go get it. Period! Forum Moderator Status: mission completed! Joined: 02 Jul 2009 Posts: 1426 GPA: 3.77 Followers: 180 Kudos [?]: 844 [1] , given: 621 Re: Counting digits (m09q17) [#permalink] ### Show Tags 17 Sep 2010, 06:46 1 This post received KUDOS 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 600-650 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. _________________ Audaces fortuna juvat! GMAT Club Premium Membership - big benefits and savings CIO Joined: 02 Oct 2007 Posts: 1218 Followers: 94 Kudos [?]: 905 [1] , given: 334 Re: Counting digits (m09q17) [#permalink] ### Show Tags 18 Sep 2010, 05:51 1 This post received KUDOS 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 big4-auditor, 300 is just a reconciliation that my counting is right. _________________ Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas. GMAT Club Premium Membership - big benefits and savings Manager Joined: 05 Jul 2008 Posts: 139 GMAT 1: Q V GMAT 2: 740 Q51 V38 Followers: 3 Kudos [?]: 103 [0], given: 40 Re: Counting digits [#permalink] ### Show Tags 07 Jul 2008, 20:11 aaron22197 wrote: I have always had trouble with "counting digits" problems I struggled with this problem during a test(it took me 3.5 mins to crack it) How many times will the digit 7 written when listing integers from 1 to 1000 ? My standard approach is- I try to figure out in how many ways can each slot be occupied 1 time between 1 and 10 19 times between 1 and 100 For 3 digit numbers ----------------------7 X Y ---|-- X Y 7 --|-- X 7 Y number of times----10 10------9 10---------9 10 -------------------------100---------90------------90 = 280 Total number of times = 280+19+1=300 So , I was happy with this above strategy and I thought I could apply it to any counting problem. Today, I encountered a fairly simple question 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? * 70 * 77 * 126 * 133 * 140 I used the simple counting method to to 19*7 = 133 However, I could not get the same answer by following the X Y Z strategy shown above. Two questions for the experts A)What am I doing wrong in the 2nd problem ? B) Is there a better approach than what I am doing? Any help appreciated... I think your approach is very good. A)What am I doing wrong in the 2nd problem ? The 1st asks "How many times will the digit 7 written?" when the 2nd asks about the number of pages that means the number of numbers containing digit 9. I think from 1-9: 1 page. From 10-99: 18 pages (X9: 8 (X≠9) 9Y: 9 (Y≠9) 99) From 100-705: (1≤ X ≤ 6) XY9: 6*9=54 pages (X, Y ≠ 9) X9Z: 6*9=54 pages (X, Z ≠ 9) X99: 6 So the number of pages is: 1+18+54+54+6=133 :D Manager Joined: 04 Apr 2008 Posts: 225 Location: Pune Followers: 2 Kudos [?]: 36 [0], given: 3 Re: Counting digits [#permalink] ### Show Tags 08 Jul 2008, 09:32 Hey Oski Verry good reasoning...............+1 for you Oski wrote: Actually you don't have to make "slices" (1-9, 10-99, 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 _________________ Every Problem Has a Sloution So keep working AB Director Joined: 25 Oct 2008 Posts: 608 Location: Kolkata,India Followers: 12 Kudos [?]: 766 [0], given: 100 Re: Counting digits (m09q17) [#permalink] ### Show Tags 19 Sep 2009, 01:00 Problem 1: numbers from 001 to 999 7 can appear as XX7,X7X,7XX where X is any value from 0-9 XX7- 10*10 ways=100 same for X7X and 7XX. total 300 ways .......... Problem 2: 001-699(lets say) 9XX >not valid for this set X9X 7*1*10=70 ways XX9 7*10*1=70 ways total 140 ways.. Whats the OA? _________________ http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902 Director Joined: 25 Oct 2008 Posts: 608 Location: Kolkata,India Followers: 12 Kudos [?]: 766 [0], given: 100 Re: Counting digits (m09q17) [#permalink] ### Show Tags 05 Oct 2009, 18:30 Guys anybody from my method??help!! So i got 140 and the answer is 133 thats means 99,199,299,399,499,599,699 have to be substracted as explained above..right?hmm..i think this is it..nice!! _________________ http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902 Intern Joined: 26 Aug 2010 Posts: 23 Followers: 0 Kudos [?]: 18 [0], given: 2 Re: Counting digits (m09q17) [#permalink] ### Show Tags 31 Aug 2010, 07:04 First of all I counted the numbers from 9 to 99 with digit 9 and the result is 19. Secondly, because we are given that the last page is 705, the hundred division could not be 9. It means that the same 19 numbers are repeated 6 more times from 100 to 699 (the last one). In total, 7*19=133. Manager Joined: 01 Apr 2010 Posts: 165 Followers: 3 Kudos [?]: 16 [0], given: 6 Re: Counting digits (m09q17) [#permalink] ### Show Tags 31 Aug 2010, 08:49 2 steps 1. 9,19,.... 99 => 10 numbers so 10*7= 70 (1-100, 100-200,.... 600-700) 2. 90-98 => 9 numbers so 9*7 =63 (1-100, 100-200,....600-700) 1+2 => 70+63= 133 D. Manager Joined: 16 Feb 2010 Posts: 225 Followers: 2 Kudos [?]: 277 [0], given: 16 Re: Counting digits (m09q17) [#permalink] ### Show Tags 31 Aug 2010, 13:35 is there a better way than the long counting method? IMO: C 133 SVP Joined: 30 Apr 2008 Posts: 1887 Location: Oklahoma City Schools: Hard Knocks Followers: 40 Kudos [?]: 564 [0], given: 32 Re: Counting digits (m09q17) [#permalink] ### Show Tags 31 Aug 2010, 13:56 Depending on what you consider the "long counting" method, yes there is another way. You know that for every 100 pages, there will be a 9 in the units once every 10 pages. So, per 100 pages, that's 10. Then for the 90's we'll need to count every page in that 10 pages, but we've already counted 99 when we counted those pages with a "9" in the units place. So we have 10 (units) + 9 (ten's place) = 19. Now we have 7 groups of 100...so 7 * 19 = 133. zisis wrote: is there a better way than the long counting method? IMO: C 133 _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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31 Aug 2010, 19:00
with a picture, this problem can be easily worked out.

Attachment:
File comment: excel grid

snap024.png [ 6.23 KiB | Viewed 7572 times ]

1) the 2nd column from the right, red letter with white background, i.e., from 9 to 89, contains nine 9s;
2) 89 makes up an extra 9;
3) the bottom line, red letter with yellow and blue background, contributes altogether ten 9s, but should be counted as nine pages as required by the problem.

so from 1~100, we count 19 (9+1+9=19) pages in the book.

following the method, there should be another 19 such pages from 101~200 pages, and so on...

and the final: 19x7=133.

done.
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01 Sep 2010, 00:48
Hello guys,
I am new here. Thanks for sending me the mail. I chose "D" . I explain like that:
1. Suppose digit "9" appears in the first site (correspond with site "a") of the numbers have two digits (was called ab), so that site "b" has 10 possibilities (0,1,2,3,4,5,6,7,8,9)
2. If digit "9" appears in the middle (correspond with site "b") of the numbers have three digits (are called "abc"), site "a" has 7 possibilities (0,1,2,3,4,5,6) (because abc<705), site "c" has 10 possibilities (0,1,2,3,4,5,6,7,8,9). Total 70 possibilities.
3. If digit "9" appears in the bottom (correspond with site "c") of the numbers have three digits (are called "abc"), site "a" has 6 possibilities (1,2,3,4,5,6), site "b" has 9 possibilities (1,2,3,4,5,6,7,8,9). Total (6*9=54) possibilities.
At last we have a same number in case 1 and 2 so that (10 +70 +54 - 1) = 133 numbers meet order of the question.
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01 Sep 2010, 00:49
Welcome to GMAT Club and good luck with your GMAT Prep!

Nguyento wrote:
Hello guys,
I am new here. Thanks for sending me the mail. I chose "D" . I explain like that:
1. Suppose digit "9" appears in the first site (correspond with site "a") of the numbers have two digits (was called ab), so that site "b" has 10 possibilities (0,1,2,3,4,5,6,7,8,9)
2. If digit "9" appears in the middle (correspond with site "b") of the numbers have three digits (are called "abc"), site "a" has 7 possibilities (0,1,2,3,4,5,6) (because abc<705), site "c" has 10 possibilities (0,1,2,3,4,5,6,7,8,9). Total 70 possibilities.
3. If digit "9" appears in the bottom (correspond with site "c") of the numbers have three digits (are called "abc"), site "a" has 6 possibilities (1,2,3,4,5,6), site "b" has 9 possibilities (1,2,3,4,5,6,7,8,9). Total (6*9=54) possibilities.
At last we have a same number in case 1 and 2 so that (10 +70 +54 - 1) = 133 numbers meet order of the question.

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01 Sep 2010, 00:57
Did I explained clearly? Please correct my wrong. Maybe my method isn't the best manner, I just want to grow rich the forum. Thanks. I want to make friends with all of you.
Re: Counting digits (m09q17)   [#permalink] 01 Sep 2010, 00:57

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# Counting digits (m09q17)

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