GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 24 Sep 2019, 02:01 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # M01-10

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 58136

### Show Tags

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.
_________________
Intern  B
Joined: 03 Sep 2017
Posts: 18
Location: Brazil
GMAT 1: 730 Q49 V41 ### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 58136

### Show Tags

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.
_________________
Intern  B
Joined: 22 Sep 2016
Posts: 46
Location: India
Concentration: International Business, General Management
GMAT 1: 680 Q44 V39 ### Show Tags

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?
_________________
Best,
Spiritual Yoda
VP  P
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)

### Show Tags

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

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 58136

### Show Tags

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

Go to page   Previous    1   2   [ 27 posts ]

Display posts from previous: Sort by

# M01-10

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  