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


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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...

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!
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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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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)

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.
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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

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?
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countdown-beginshas-ended-85483-40.html#p649902

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.

is there a better way than the long counting method?
IMO: C 133

with a picture, this problem can be easily worked out.



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.

Welcome to GMAT Club and good luck with your GMAT Prep!


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