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pages of the book

by uptowngirl92 » Sat Sep 19, 2009 1:02 am
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

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..

where am i going wrong??
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by aim-wsc » Sat Sep 19, 2009 2:04 am
You cannot count page# 699 twice even if it has nine appeared two times. It's just a one page right?
So you should minus the no. of pages which have nine number twice.
HTH. :)
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by gmatdriller » Mon Oct 04, 2010 7:33 am
yup, numbers of double-digit 9's are:
099
199
299
399
499
599
699---> 7 in all

140 - 7 = 133
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by Rahul@gurome » Mon Oct 04, 2010 11:02 pm
Solution:
The digit 9 will appear on the pages 9, 19, 29, 39......,90,91,....99,109,...190, 191,...... 199,209,...... 290, 291,...... 299,309....390,391, ......399,409.....490, 491,......499,509.......590,591,......,599, 609.......690, 691,.....,699.

Or 9 appears on pages 9, 19, 29, .....89, 99, 109,........699 and also on pages 90,91,...98,...., 190, 191,....198,.......290, 291,......298,.......390, 391, .........398,.....490, 491,....498,....590, 591,..598,....,690, 691,....698,.

Now 9, 19, 29, .....89, 99, 109,........699 forms an arithmetic progression with number of elements =( 699 - 9)/10 + 1 = 70.

Also the sequence 90,91,...98,...., 190, 191,....198,.......290, 291,......298,.......390, 391, .........398,.....490, 491,....498,....590, 591,..598,....,690, 691,....698, will have 9*7 = 63 elements.

So the number of pages on which 9 appears is (699 - 9)/10 + 1 + 9*7 = 70 + 63 = 133.
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by gmatdriller » Wed Nov 10, 2010 1:44 am
@Rahul
don't you think the approach you used is longer than that above?
Am only considering the 2min limit.
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by thebigkats » Thu Dec 23, 2010 11:09 am
Hi My approach took a total of 30 sec:
I created a quick column for 1-100 and wrote down the no's
then on top wrote other 100bases leading to 705 (100, 200, 300, 400, 500, 600)

1-100 lead me to 19 pages with 9's and multiply it with 7 gives me 133

I could have used fancy formulas but it justs seemed easy given that the last page ended squarely at 705 (no 9's after 700)
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by Zerks87 » Thu Dec 30, 2010 12:37 pm
gmatdriller wrote:@Rahul
don't you think the approach you used is longer than that above?
Am only considering the 2min limit.
This is a quicker way to do the question. This took me maybe a minute.

The trick with this question (and I fell into it, is to forget to count 90-99 for every 100 pages)

In the first 100 pgs you have:

9, 19, 29, 39, 49, 59, 69, 79, 89 and 90-99 or ((99-90)+ 1 items) = 10

So for every 100 pgs you have 9 + 10 pages where there is a nine.

Since you have pages 100-200, 200-300, 300-400, 400-500, 500-600, 600-700

The total number of times a 9 appears in the pages = 19 (for the first 100 pages) + 19 (6) = 133
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by saleem.kh » Thu Apr 07, 2011 5:14 am
Yes. I noticed that Rahul often uses indirect method, anyway thanks to him for solving many problems and explaining them on queries.

This problem is simply to calculate number of pages with "9" in first 100 pages... i.e. 19 and then multiply it with 7 = 133.
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by GMATGuruNY » Thu Apr 07, 2011 7:19 am
uptowngirl92 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?

* 70
* 77
* 126
* 133
* 140

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..

where am i going wrong??
We could treat this as an overlapping groups question:

Total = (Integers with a tens digit of 9) + (Integers with a units digit of 9) - (Integers with both a tens digit of 9 and a units digit of 9).

In the solution below, a hundreds digit of 0 represents a 1-digit or 2-digit integer.

Tens digit of 9:
Hundreds digit can be 0 through 6 = 7 choices.
Units digit can be 0 through 9 = 10 choices.
Multiplying, we get 7*10 = 70 integers.

Units digit of 9:
Hundreds digit can be 0 through 6 = 7 choices.
Tens digit can be 0 through 9 = 10 choices.
Multiplying, we get 7*10 = 70 integers.

Both a tens digit of 9 and a units digit of 9:
Hundreds digit can be 0 through 6 = 7 choices.

Total = 70 + 70 - 7 = 133.

The correct answer is D.
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by nafiul9090 » Fri Apr 08, 2011 9:25 pm
0-100, the total number of 9 is 19
So total number of 9 from 0-699 is 19*7=133
But 7 pages each has extra 9
So in my point of view, 9 appears in 133-7=126 pages
Correct me if I am wrong
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by thebigkats » Fri Apr 08, 2011 9:44 pm
Hi:
Lets keep it simple... no of pages from 1-100 -
9 , 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99 (TOTAL 19)

same pattern repeats for 100-200, 201-300, 301-400, 401-500, 501-600 and 601-700 (TOTAL 6 more such 'groups')

TOTAL = 19 * 7 = 133

regards,
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by OneTwoThreeFour » Fri Apr 08, 2011 10:00 pm
The best way to solve this is to avoid the double counting in the first place.

7*1*9=63
7*9*1=63

Now include the double 99s: 099-699. (There are seven of them)

Thus 63+63+7= 133
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by edvhou812 » Mon Apr 11, 2011 10:40 pm
First one must ask, "Will this be really be used in real life?" Then you get to solve it anyway. ;)

9 appears nine times in each set of two digits. Example: 9, 19, 29 all the way up to 89. 9 then appears 10 times from 90-99. So it appears 19 times from 1-99, and then repeats once we get to 100-199, etc. We go up to 705, so forget about the 700s. We are left with 19x7=133.
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by parul9 » Sat Nov 05, 2011 8:03 am
1-10 -> 1
11-20 -> 1
21-30 -> 1
31-40 -> 1
41-50 -> 1
51-60 -> 1
61-70 -> 1
71-80 -> 1
81-90 -> 2(89 and 90)
91-100-> 9

Total no. of 9s from 1 to 100 = 19

so, total 9st from 1 to 700 will be 19 * 7 = 133.
From 701 to 705, there is no occurance of 9.

So, answer would be 133.
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by ArunangsuSahu » Tue Jan 03, 2012 8:57 pm
Let's make it simpler

1) 0-99 has 20 9s.
2) 100 to 699 has 20*6=120 9s.
So total 9s = 20+120=140

Now here the number of pages are wanted.

99-699=7 pages have double 9s

so total no of pages with numeration as 9 = 140-7=133
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