Hello Bunuel

Could you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000?
I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused.

Thank you,

Bunuel chetan2u

The answer will be 300 for all the digits except 0 and 1, right? For 0, the answer will be 303 (Considering 3 zeroes in 1000) and for 1, the answer will be 301 (Considering one 1 in 1000).

Let me know if my understanding is correct?

Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304


Many approaches are possible. For example:

Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of \(3*1000=3000\) digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) \(\frac{3000}{10}=300\) times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the tens digit or the units digit. Case 1: 7 is the tens digit. There are 10 ways to place 7 as the tens digit of a two-digit number. Case 2: There are 9 ways to place the units digit. Thus, for two-digit numbers we have: \(10+9=19\) numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

In the range 0-100:

7 as units digit - 10 times (7, 17, 27, ..., 97);

7 as tens digit - 10 time (71, 72, 73, ..., 79);

So in first one hundred numbers 7 is written \(10+10=20\) times.

In 10 hundreds 7 as units or tens digit will be written \(10*20=200\) times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total \(200+100=300\).


Answer: D

Hi Bunuel,

In approach # 1, i didn't quite understand why did we use 3 digit numbers.
Can you please explain?
Apologies if it's a silly question.

Official Solution:


Approach #1:

Consider numbers from 0 to 999 written as follows:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of \(3*1000=3000\) digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) \(\frac{3000}{10}=300\) times.

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304

This is how I did..same approach as you took...this might help you understand what went wrong

_ single digit -> 1 possibility
_ _ two digits -> (1 possibility * 10 possibilities) + (9 possibilities * 1 possibility) -> (9, as we cannot place a 0) = 19
_ _ _ three digits -> (1 possibility * 10 possibilities * 10 possibilities) + (9 possibilities * 1 possibility * 10 possibilities) + (9 possibilities * 10 possibilities * 1 possibility) = 100 + 90 + 90 = 280

add up, 1 + 19 + 280 -> 300

Hi..
if we write all as 000 to 999, we will have 3000 digits as 3(number of digits in each number )*1000(total numbers)=3000..
In these 3000, all 10 digits will be 3000/10=300..

But we do not have 0s in single digit and two digits..
so 000,as we do not have 0, so 3 0s missing
00a.. from 1 to 9, they are not written as 001,002...009, so 2*9=18 x 0s missing
0ab.....from 10 to 99, they are not written as 010,011...099, so 1*90=90 x 0s missing
Total 3+18+90=111 number of 0s are less..

Thus, from 1 to 999, we have 300 of all 9 digits - 1 to 9, while we have 300-111=189 of 0s.
Add 1000,
digit 0 is used 189+3=192 times
digit 1 is used 300+1=301 times
digits 2 to 9, 300 times each

@Bunuel/ chetan2u

As mentioned in earlier thread that there is no preference of one digit over another, for 0s and 1 also be 300. (how can it be 192 and 301 respectively?) Plz shed some lights.

possibilities are

X X 7 = 9*9*1
X 7 X = 9*1*9
7 X X = 1*9*9
7 7 X = 1*1*9
7 X 7 = 1*9*1
X 7 7 = 9*1*1
777 = 1
if we add all of them = 81+81+81+9+9+9+1=271
where did i go wrong??

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304