Around the World in 80 Questions (Day 10): A professor is numbering

Question

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

Oppenheimer1945 · Answer

Zeros between 1 to 1000

1 to 100 = 11
101 to 200 = 20
201 to 300 = 20
301 to 400 = 20
401 to 500 = 20
501 to 600 = 20
601 to 700 = 20
701 to 800 = 20
801 to 900 = 20
901 to 1000 = 21

Total - 192 zeros.
Hence C

snoep · Answer

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'?

10, 20, 30, 40, 50, 60, 70 , 80 , 90 = 9
100, 200, 300, 400, 500, 600, 700 , 800 , 900 = 18
101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 130, 140, 150, 160, 170, 180, 190 = 18
201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 220, 230, 240, 250, 260, 270, 280, 290 = 18
301........................................................................ = 18
401........................................................................ = 18
501........................................................................ = 18
601........................................................................ = 18
701........................................................................ = 18
801........................................................................ = 18
901........................................................................ = 18
1000 = 3

9 + 18*10 + 3 = 192 
 Answer: C

missionmba2025 · Answer

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

vishal_gupta · Answer

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.

sujoykrdatta · Answer

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

pintukr · Answer

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

mialanknox · Answer

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):   = 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.

XLmafia21 · Answer

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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