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How many times will the digit 7 be written when listing the integers 11 Sep 2018, 17:21
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seekmba
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
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The digit 7 as the hundreds digit appears 100 times (700 to 799). As the tens digit, it appears 100 times also (ten times each in the 70s, 170s, 270s, …, 970s). As the units digit, it appears 100 times also (7, 17, 27, …, 997). Therefore, the digit 7 has appeared a total of 300 times.

Answer: D
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How many times will the digit 7 be written when listing the integers 12 Dec 2020, 11:49
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I did by total combinations- restricted (no 7's).

Total combination=10*10*10=1000
no 7's in the digit= 9*9*9=729

no of times 7 will be written from 1-1000= 1000-729=271. But the answer is 300. What mistake did i do here?
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How many times will the digit 7 be written when listing the integers 12 Dec 2020, 21:55
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Hi Zeus666,

Your approach calculated the number of integers that had AT LEAST ONE '7' in the digits. However, that's not what the question asked for. We need to total all of the 7's that appear in the first 1,000 positive integers - and your solution does not consider numbers that include more than one 7 as digits (for example, 717, 778 or 777).

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How many times will the digit 7 be written when listing the integers 22 May 2021, 04:19
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Solution:

Total 3 digit numbers from 1-1000,if we represent each number as abc (where a and b are 0 for double and single digit numbers respectively)
=3*1000 = 3000

Every digit must be equally used and hence each digit is used 3000/10 =300 times
=>7 shall be used as a digit 300 times too. (option d)

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How many times will the digit 7 be written when listing the integers 10 May 2023, 20:43
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KarishmaB What is wrong with below approach?
XX7 -> This way 100 numbers are possible.
X7X -> This way 100 numbers are possible.
7XX -> This way 100 numbers are possible.
Total 300. But there will few same number such as 777 (counted thrice), 77 (counted twice). So we will have to subtract it.
Answer will come 297
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How many times will the digit 7 be written when listing the integers 10 May 2023, 22:52
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saketkandoi
KarishmaB What is wrong with below approach?
XX7 -> This way 100 numbers are possible.
X7X -> This way 100 numbers are possible.
7XX -> This way 100 numbers are possible.
Total 300. But there will few same number such as 777 (counted thrice), 77 (counted twice). So we will have to subtract it.
Answer will come 297
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There are two different concepts:

How many times is the number 7 written? You will write 7 three times in 777 so it will be counted 3 times. You are not to remove 2 out of it.

How many numbers will have 7 in them? Now 777 is only one such number so you will count it only once.

Since our question asks us to find how many times 7 was written, it was written 100 times in hundreds place, 100 times in tens place and 100 times in units place so total 300 times.
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How many times will the digit 7 be written when listing the integers 04 Jun 2025, 12:27
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Are there more problems like this?
Bunuel? Thanks
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How many times will the digit 7 be written when listing the integers 04 Jun 2025, 12:32
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Are there more problems like this?
Bunuel? Thanks
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Check below:
https://gmatclub.com/forum/how-many-tim ... 23406.html
https://gmatclub.com/forum/the-digit-7- ... 24870.html
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How many times will the digit 7 be written when listing the integers 24 Oct 2025, 05:57
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in approach 1 why did you start with three digit like i didnt understand that logic?
Bunuel
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


Multiple approaches are possible to solve this problem. Here are two:

Approach #1:

Let's consider the numbers from 0 to 999, which are written using three digits:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

There are a total of 1000 numbers, each of which uses 3 digits. Therefore, there are a total of \(3*1000=3000\) digits used in all these numbers. Since there is no reason for any digit to be favored over another, each of the 10 digits should be used an equal number of times. Hence, each digit (including 7) is used \(\frac{3000}{10}=300\) times.

Approach #2:

Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). Therefore, in the first 100 numbers, the digit 7 is written a total of 10 + 10 = 20 times.

Within each block of 100 numbers (e.g., 1-100, 101-200, ..., 901-1000), the digit 7 appears 20 times as the units or tens digit. Hence, it appears as the units or tens digit 20*10 = 200 times in the 10 blocks. Additionally, the digit 7 appears 100 times as the hundreds digit (700, 701, 702, ..., 799).

Thus, the digit 7 is written a total of 200 + 100 = 300 times in the range of 0-999.


Answer: D
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How many times will the digit 7 be written when listing the integers 24 Oct 2025, 06:04
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sakshijjw
in approach 1 why did you start with three digit like i didnt understand that logic?

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We start with three-digit numbers (000–999) because it keeps all numbers the same length, making it easy to count how often each digit appears in each place. It’s just a counting trick, not because the numbers themselves have three digits.
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How many times will the digit 7 be written when listing the integers 24 Oct 2025, 06:20
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so if it were 10000 instead of 1000 i will start the series with 0000 so the answer would be 10000*4/10 = 4000

How will you solve 10000 using Approach 2 --- because i tried and i got stuck at 20*100= 2000 +100+ 1000 = 3100

PLEASE HELP
Bunuel

We start with three-digit numbers (000–999) because it keeps all numbers the same length, making it easy to count how often each digit appears in each place. It’s just a counting trick, not because the numbers themselves have three digits.
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How many times will the digit 7 be written when listing the integers 24 Oct 2025, 06:37
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sakshijjw
so if it were 10000 instead of 1000 i will start the series with 0000 so the answer would be 10000*4/10 = 4000

How will you solve 10000 using Approach 2 --- because i tried and i got stuck at 20*100= 2000 +100+ 1000 = 3100

PLEASE HELP

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Yes, for 0 - 9999 you can use 0000 - 9999. Each of the 4 positions takes digit 7 exactly 10000/10 = 1000 times, so total = 4*1000 = 4000.

Approach 2:

Units + tens: 20 per 100-block, and there are 100 blocks, so 20*100 = 2000.

Hundreds: 100 per 1000-block, and there are 10 such blocks, so 100*10 = 1000 (you used 100 by mistake).

Thousands: 7000 - 7999, so 1000.

Total = 2000 + 1000 + 1000 = 4000.
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How many times will the digit 7 be written when listing the integers 24 Oct 2025, 10:08
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I used an approach that I actually don't know why it worked.

I counted the ammount of times that 7 appeared in each frequency:

1x:

1 . 9 . 9 (3p1)

2x:

1 . 1 . 9 (3p2) (x2 to get number of seven's)

3x:

1 . 1 . 1 (x3 to get number of seven's)

=300

However, why did it worked if I didn't excluded the 0 in the hundreds (for example: 8 . 9 . 1 . (3p1) in the 1x)?
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How many times will the digit 7 be written when listing the integers 11 Nov 2025, 16:32
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Quick question, why is it not double counting in approach 2 when you say "Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). Therefore, in the first 100 numbers, the digit 7 is written a total of 10 + 10 = 20 times" ?

You are considering 77 when the 7 is in the units digit but also considering 77 when 7 is in the tens digit, right?
Bunuel
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


Multiple approaches are possible to solve this problem. Here are two:

Approach #1:

Let's consider the numbers from 0 to 999, which are written using three digits:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

There are a total of 1000 numbers, each of which uses 3 digits. Therefore, there are a total of \(3*1000=3000\) digits used in all these numbers. Since there is no reason for any digit to be favored over another, each of the 10 digits should be used an equal number of times. Hence, each digit (including 7) is used \(\frac{3000}{10}=300\) times.

Approach #2:

Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). Therefore, in the first 100 numbers, the digit 7 is written a total of 10 + 10 = 20 times.

Within each block of 100 numbers (e.g., 1-100, 101-200, ..., 901-1000), the digit 7 appears 20 times as the units or tens digit. Hence, it appears as the units or tens digit 20*10 = 200 times in the 10 blocks. Additionally, the digit 7 appears 100 times as the hundreds digit (700, 701, 702, ..., 799).

Thus, the digit 7 is written a total of 200 + 100 = 300 times in the range of 0-999.


Answer: D
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How many times will the digit 7 be written when listing the integers 11 Nov 2025, 20:37
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True I made same mistake too how do u improve at problem solving for gmat focus edition if you can help
pradosamant
in case of counting from 0 to 100 you are missing 77--the count is 2 here for 7, you considered just one 7.
So for all 0 to 1000 you are missing total of 10 such counts.
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How many times will the digit 7 be written when listing the integers 12 Nov 2025, 00:00
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Alejocz92
Quick question, why is it not double counting in approach 2 when you say "Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). Therefore, in the first 100 numbers, the digit 7 is written a total of 10 + 10 = 20 times" ?

You are considering 77 when the 7 is in the units digit but also considering 77 when 7 is in the tens digit, right?

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77 is counted twice because it contains two 7s, one in the tens place and one in the units place, so both must be included. This is explained many times in the thread, Please review.
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