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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?

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?

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$$.

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.

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

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

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

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? _________________

Re: Counting digits (m09q17) [#permalink]
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!! _________________

Re: Counting digits (m09q17) [#permalink]
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.

Re: Counting digits (m09q17) [#permalink]
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$$.

Re: Counting digits (m09q17) [#permalink]
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 6807 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...

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.

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.

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.

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...

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...

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.