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# M01-10

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Math Expert
Joined: 02 Sep 2009
Posts: 58136

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24 Dec 2017, 01:37
Siddhuftr wrote:
ones - x - 1 time
tens - x7 + 7x - 9*1 + 1*10 = 19 times
Hundreds - xx7 + x7x + 7xx = 9*10*1 + 9*1*10 + 1*10*10 = 190
Total = 300

I am unable to understand , how double counting of 77 , 777 is avoided ?

The question asks "How many times will the digit 7 be written when listing the integers from 1 to 1000?" not how many number will have 7. So, for example, for number 777, it should be counted as three 7's not one.
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Intern
Joined: 03 Sep 2017
Posts: 18
Location: Brazil
GMAT 1: 730 Q49 V41

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21 May 2018, 09:25
Bunuel wrote:
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.

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,
Math Expert
Joined: 02 Sep 2009
Posts: 58136

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21 May 2018, 21:28
vitorcbarbieri wrote:
Bunuel wrote:
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.

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,

Since neither 0 not 1000 has 7 in it, then excluding/including these values into the range won't change the answer.
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Intern
Joined: 22 Sep 2016
Posts: 46
Location: India
Concentration: International Business, General Management
GMAT 1: 680 Q44 V39

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03 May 2019, 22:15
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?
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Spiritual Yoda
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Joined: 14 Feb 2017
Posts: 1052
Location: Australia
Concentration: Technology, Strategy
Schools: LBS '22
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
WE: Management Consulting (Consulting)

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29 May 2019, 16:25
Hey Bunuel, what is meant by "In 10 hundreds 7" in approach 3? I'm having a hard time understanding this.
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Goal: Q49, V41
Intern
Joined: 22 Dec 2018
Posts: 17
Concentration: Healthcare, International Business
WE: Medicine and Health (Health Care)

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11 Sep 2019, 11:09
Bunuel wrote:
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 $$200+100=300$$.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58136

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17 Sep 2019, 05:54
swatjazz wrote:
Bunuel wrote:
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 $$200+100=300$$.

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.

Not a silly question at all.

Let me try to elaborate: this approach worked because when we write the numbers from 0 to 999 in the form XXX each digit take the values from 0 to 9 which provides that in the end all digits are used equal # of times.

For the range 100 to 999 it won't be so. We can solve for this range in the following way:
XX7 - 7 in the units place - first digit can take 9 values (from 1 to 9) and second digit can take 10 values (from 0 to 9) --> total numbers with 7 in the units place: 9*10=90;

X7X - 7 in the tens place - first digit can take 9 values (from 1 to 9) and third digit can take 10 values (from 0 to 9) --> total numbers with 7 in the tens place: 9*10=90;

7XX - 7 in the hundreds place - second digit can take 10 values (from 0 to 9) and third digit can take 10 values (from 0 to 9) --> total numbers with 7 in the hundreds place: 10*10=100.

TOTAL: 90+90+100=280.

Hope it helps.
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Re: M01-10   [#permalink] 17 Sep 2019, 05:54

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# M01-10

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