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

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

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

- '0' appears once in numbers from 10 to 19. This pattern repeats for every 100 (e.g., 100-109, 200-209, etc.). So, from 1 to 1000, this gives us 10 * 10 = 100 occurrences.
- '0' appears once in every 10 numbers (e.g., 20, 30, 40, etc.). This pattern repeats for every 100. So, from 1 to 1000, this gives us 10 * 10 = 100 occurrences.
- However, we've double-counted the '0's in numbers like 100, 200, 300, etc. So, we subtract these 10 occurrences.
- Finally, we add the '0's in 1000, which gives us 2.
So, the total is 100 + 100 - 10 + 2 = 192. Therefore, the correct answer is C. 192.

number of zeroes from 1 to 100 are 11
zeroes from 101 to 109 are 9
so total zeroes from 101 to 200 are 20
101 to 900 total zeroes will be 160
and from 901 to 1000 ; 20+1 ; 21
11+160+21 ; 192
OPTION C

There are total 9 numbers between 1 to 99 which have only one zero are :
10,20,30,40,50,60,70,80,90
Therefore, total one zero numbers between 1 and 999( x0 format) are 9*10= 90 zeros.
And zeros which come in between 3 digit numbers are: eg: 101,102,103….109
Therefore, total one zero numbers between 1 and 999(x0x format) are 9*9= 81 zeros.
Now, 2 zero numbers in between 1 and 999 are:
100,200….900
So , No. Of zeros (x00 format)= 2*9 = 18
And 3 zeros of 1000 .
Therefore, total zero's are : 3+18+81+90 = 192.

Posted from my mobile device

1-digit numbers: no zero
2-digit numbers: 9 zeros
3-digit numbers: from 100 to 999, there are 900 numbers. Of those 900 numbers, 0 can only be the tenth or digit numbers -> There are 900/10*2=180 zeros
There are 3 zeros in number 1000

-> Total zeros = 9+180+3=192
Answer C

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'?
_______________________
for 10th such as 10, 20 , 30,...,110,120 ... 990 - 9X9+9=90
for 100th such as 100,200, 300 .. 900 - 2x9=18
for 1000 - 3
from 101 to 109, 201 to 209 ... - 9x9=81

90+18+3+81=193
D is the winner.

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

1 to 1000 with how many 0s professor will be writing
1 to 100 will have 10, 20, 30, 40, 50, 60 , 70 , 80, 90 = 9
100 to 199 will have 100= 2 times, 101 to 109 = 9 times , 110 to 190 = 9 times; Total = 20
Similarly will be the case from 200 to 900
and 1000 will have 3 zeros
so total zeros = 9 +20*9+3= 192 zeros

C is the 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.

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

10,20....90 - 9 zeros
101,102,103,...109 - 9 zeros
110, 120,130,...190 - 9 zeros
201,202,203,209 - 9 zeros
210, 220,230,...290 - 9 zeros
301,302,303,....309 - 9 zeros
310, 320,330,...390 - 9 zeros
401,402,403,....409 - 9 zeros
410, 420,430,...490 - 9 zeros
501,502,503,....509 - 9 zeros
510, 520,530,...590 - 9 zeros
601,602,603,...1609 - 9 zeros
610, 620,630,...690 - 9 zeros
701,702,703,...709 - 9 zeros
710, 720,730,...190 - 9 zeros
801,802,803,...809 - 9 zeros
810, 820,830,...890 - 9 zeros
901,902,903,...909 - 9 zeros
910, 920,930,...990 - 9 zeros

100,200,300,...900 - 9*2 = 18 zeros

1000 = 3 zeros

Total zeros used = 9*19 + 18 + 3 = 192

IMO C

Case 1: Numbers from 10->90. 9 Zeros
Case 2: Numbers like 100, 200, 300 till 900. 9*2 = 18 zeros
Case 3: Numbers like 101, 102, 103.. 9*9 = 81 zeros
Case 4: Numbers like 110, 120, 130.. 9*9 = 81 zeros
Case 5: The 3 zeros in 1000 also to be included.

Sum it all up and you get 192 zeros.

the Answer is C
The Professor will write 100 times 0 in the ones place
He will write 10 times 0 in the hundreds place
He will write 1 times 0 in 1000 place
He will write 9 times 0 from 100-200 to include 110,120...190) So 9 x 9 for 100-900.
Hence the total number of zeroes come out as 100+10+1+81=192

There are 9 numbers between 1 and 99 that have one zero: 10, 20, 30, 40, 50, 60, 70, 80, and 90.

This means that there are 90 numbers between 1 and 999 that have one zero in the x0 format.

For example, 101, 102, 103, and so on.

This means that there are 81 numbers between 1 and 999 that have one zero in the x0x format.

For example, 101, 110, 120, and so on.

There are also 18 numbers between 1 and 999 that have two zeros in the x00 format.

For example, 100, 200, 300, and so on.

Finally, there are 3 zeros in the number 1000.

Therefore, there are a total of 3 + 18 + 81 + 90 = 192 zeros between 1 and 999.

The digit '0' will be written 192 times.

There are 9 numbers in the form "d0", where d is one of the digits 1-9.
There are 9 numbers in the form "d00", where d is one of the digits 1-9.
There are 81 numbers in the form "d0k", where d and k are one of the digits 1-9.
There are 81 numbers in the form "dk0", where d and k are one of the digits 1-9.
There is 1 number in the form "1000".
In total, this is 9 + 9 + 81 + 81 + 1 = 192 times.

Answer is C
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


from 10,20 to 90 we have 9
from 100 to 900 we have 9*2=18
1000 we have 3
from 101 to 909 we have 9*9=81
from 110 to 990 we have another 81
Thus total we have 9+18+3+81+81 = 192

No. of the form X0 = 10,20...90 = 9
No. of the form 1X0 = 110, 120...190 = 9 , similarly for 2X0 ...9X0, in total = 9*9 = 81
No. of the form 10X = 101,102...109 = 9, similarly for 20X...90X, in total = 9*9 = 81.
No. of the form X00 = 100,200,...900 = 9, no. of zeros = 9*2 = 18
No. of the form 1000 = 3
So total zeros = 192.

IMO C.

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