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Bunuel
A professor is numbering the pages in a book from 1 to 1,000, inclusive (1, 2, 3, ..., 1,000). How many times will he write the digit '0'?

A. 190
B. 191
C. 192
D. 193
E. 300


 


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Single Digit Number = 0

Two Digit Number = 9 (10 / 20 / 30 / 40 / .... 90)

Three Digit Number = 100 to 199 -- 0 repeats 10 times between 100 and 109, inclusive and 10 times for 110 / 120/ 130 .... 190

In each block of 100 numbers, 0 repeats 20 times.

Between 100 and 999, we have 9 such blocks.

20 * 9 = 180

Four Digit Number: 1000, so three times

Sum = 180 + 3 + 9 = 192

IMO C
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Bunuel
A professor is numbering the pages in a book from 1 to 1,000, inclusive (1, 2, 3, ..., 1,000). How many times will he write the digit '0'?

A. 190
B. 191
C. 192
D. 193
E. 300


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

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10 - 90 = 9 zeroes
Multiple of 100s till 900 = 9*2 = 18 zeroes
Numbers after 100 with last digit zero excluding multiple of 100s = 9*9 = 81 zeroes
Numbers after 100 with middle digit zero excluding multiple of 100s and numbers with last digit zero = 9*9 = 81 zeroes = 9*9 = 81 zeroes
1000 = 3 zeroes.

So, total zeroes = 192

C is correct choice.
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Bunuel
A professor is numbering the pages in a book from 1 to 1,000, inclusive (1, 2, 3, ..., 1,000). How many times will he write the digit '0'?
A. 190
B. 191
C. 192
D. 193
E. 300
 


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The cases are:
For single digits: none are 0

For 2-digit numbers: There are 9 possibilities (10, 20 ... 90)

For 3-digit numbers:
a) Units place is 0: Number of times = 9*10 = 90
b) Tens place is 0: Number of times = 9*10 = 90

For 4 digit numbers: 1000 => 3 times

Total: 9 + 90 + 90 + 3 = 192

Answer C
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A professor is numbering the pages in a book from 1 to 1,000, inclusive (1, 2, 3, ..., 1,000). How many times will he write the digit '0'?

A. 190
B. 191
C. 192
D. 193
E. 300


For, 1-digit numbers (1 to 9 ) , Count of writing the digit '0' = 0.................(1)

For, 2-digit numbers (1 to 9 ) , Count of writing the digit '0'

10, 20., ........,90
Count of writing the digit '0' = 9.............(2)

For, 3-digit numbers (1 to 9 ) , Count of writing the digit '0'

Hundred's place can't be 0 (Hundred's place can take 9 values)

If there are 2 0's , then there are 9 possibilities (100, 200,...900) = 9 * 2 = 18 times
If there is 1 0's , then putting o at ten's place, there are 9 possibilities for unit's place = 9 * 9 = 81 times
If there is 1 0's , then putting o at unit's place, there are 9 possibilities for ten's's place = 9 * 9 = 81 times


So, Count of writing the digit '0' = 18 + 61 + 81 = 180.............(3)

For, 4-digit numbers (1 to 9 ) , Count of writing the digit '0' = 3 (ie, for 1000).................(4)

Using (1), (2), (3) and (4), we get

Total times = 0 + 9 + 180 + 3 = 192

(C) is the CORRECT answer
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Bunuel
A professor is numbering the pages in a book from 1 to 1,000, inclusive (1, 2, 3, ..., 1,000). How many times will he write the digit '0'?

A. 190
B. 191
C. 192
D. 193
E. 300


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

Win over $20,000 in prizes: Courses, Tests & more

 



The range from 1 to 1000 can be divided into 4 parts:
(1) 1 zero: 10-90
(2) 1/2 zeros: 100-990
(3) 1 zero: Set of 9 ranges from 101-109, 201-209,....901-909
(4) 3 zeros: 1000

Total numbers in range (1): (90-10)/10+1 = 9 numbers
Total zeros in range (1): 9*1 zero = 9 zeros........(A)

Total numbers in range (2): (990-100)/10+1 = 90 numbers
Total zeros in range (2): 9*2 zeros + 81*1 zero = 99 zeros........(B)

Total numbers in range (3): [Set 1: 101...109, Set 2: 201...209,....Set 9: 901-909] = 9 Sets * 9 numbers per set = 81 numbers
Total zeros in range (3): 81*1 zero = 81 zeros........(C)

Total numbers in range (4): 1000 = 1 number
Total zeros in range (4): 1*3 zeros = 3 zeros........(D)

Total zeros in (1), (2), (3), & (4): (A) + (B) + (C) + (D) = 9+99+81+3 = 192 zeros

Therefore, option C is the correct answer.
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Cases: listing down all possibilities where a 0 could occur in the 3 digits exclusively. Which means remaining places will be non-zero to make all the scenarios exhaustive and exclusive
1. _ _0 = 9*9*1 = 81 times 0 is written; here 1 is multiplied to count the number of times 0 is written
2. _0_ = 9*9*1 = 81 times 0 is written
3. 0_ _ = invalid case as units and ten's digit will be non-zero, so essentially it will be a 2-digit number with no 0
4. _00 = 9*2 = 18 times 0 is written
5. 0_0 = 9*1 = 9 times 0 is written
6. 00_ = invalid case as this is a one digit number with no 0
7. 000 = only possible case here is 1000 with 3 times 0 written
So, sum = 81 + 81 + 18 + 9 + 3 = 192 (option C)
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