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sherxon
Joined: 28 May 2021
Last visit: 09 Nov 2025
Posts: 130
Given Kudos: 30
Location: Uzbekistan
Concentration: Strategy, General Management
Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q50 V46 (Online)
GPA: 4
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Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 2: 770 Q50 V46 (Online)
Posts: 130
02 Jun 2021, 21:58
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:43) correct
40%
(02:51)
wrong
based on 251
sessions
History
Date
Time
Result
Not Attempted Yet
The total number of times that digits (0-9) were used for numbering the pages of a book is 1012. If the numbering started at page 3, how many pages are there in the book?
A. 374
B. 400
C. 506
D. 421
E. 434
A. 374
B. 400
C. 506
D. 421
E. 434
02 Jun 2021, 23:04
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sherxon
The page numbers would start from 1-digit number and finish with 3-digit numbers.
Let us find the digits used in
1) 1-digit page numbers: 3 to 9, so 7 in number. Digits used - 7
2) 2-digit page numbers: 10-99, so 90 in number. Digits used - 2*90=180
Thus, digits used in 3-digit numbers = 1012-180-7=825.
Total pages with 3-digit number= \(\frac{825}{3}=275\)...Each page will have 3 digits on it.
Total pages on the book = 99+275=374
A
02 Jun 2021, 23:04
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Instead of discussing the proper method here, I am going to write the fastest method to achieve the result:
So we know, all numbers are made up of digits from 0-9 so no need to worry about miscounting we just need to count the digits in all the number to arrive at 1012
Given: page numbering starts from 3,
We know that for a single-digit number the number of time the digits 0-9 occur is 1, so counting from 3 to 9 there are 7 page-numbers
so 7*1 = 7 -> (1)
Similarly, for 2 digit numbers there can 90 page-numbers
so 90*2 = 180 -> (2)
Now seeing the options we can see that no option is a four-digit or two-digit page number, so all two-digit numbers are possible.
Now, removing (1) & (2) from 1012 will get us the number of times digits are used in 3-digit numbers
i.e. 1012 - 7 - 180 = 825 -> (3)
Now if there are x three-digit numbers then there will be 3x digits used
Therefore 3x = 825
x = 275
Counting from 100, 374 is the 275th number
Therefore the correct option is (A)
So we know, all numbers are made up of digits from 0-9 so no need to worry about miscounting we just need to count the digits in all the number to arrive at 1012
Given: page numbering starts from 3,
We know that for a single-digit number the number of time the digits 0-9 occur is 1, so counting from 3 to 9 there are 7 page-numbers
so 7*1 = 7 -> (1)
Similarly, for 2 digit numbers there can 90 page-numbers
so 90*2 = 180 -> (2)
Now seeing the options we can see that no option is a four-digit or two-digit page number, so all two-digit numbers are possible.
Now, removing (1) & (2) from 1012 will get us the number of times digits are used in 3-digit numbers
i.e. 1012 - 7 - 180 = 825 -> (3)
Now if there are x three-digit numbers then there will be 3x digits used
Therefore 3x = 825
x = 275
Counting from 100, 374 is the 275th number
Therefore the correct option is (A)
