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

 It is currently 14 Oct 2019, 09:21 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.  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: 58318

Show Tags

6
44 00:00

Difficulty:   95% (hard)

Question Stats: 41% (02:02) correct 59% (02:21) wrong based on 300 sessions

HideShow timer Statistics

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 58318

Show Tags

6
7
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 $$200+100=300$$.

_________________
Intern  Joined: 08 Aug 2014
Posts: 7

Show Tags

Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

I make the answer C.
Math Expert V
Joined: 02 Sep 2009
Posts: 58318

Show Tags

hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

I make the answer C.

7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.
_________________
Retired Moderator B
Status: Getting strong now, I'm so strong now!!!
Affiliations: National Institute of Technology, Durgapur
Joined: 04 Jun 2013
Posts: 390
Location: United States (DE)
GPA: 3.32
WE: Information Technology (Health Care)

Show Tags

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??
Manager  Joined: 24 Nov 2013
Posts: 56

Show Tags

3
1
WaterFlowsUp wrote:
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??

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
Intern  Joined: 16 Aug 2013
Posts: 11

Show Tags

Bunuel wrote:
hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

I make the answer C.

7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.

Hi Bunuel
I understand your logic, but what if we count the numbers without 7 such that we have a 3 digit number that can be made by numbers {0,1,2,3,4,5,6,8,9} and then subtract the result from 1000. given this logic, 1000-9*9*9= 271
Math Expert V
Joined: 02 Sep 2009
Posts: 58318

Show Tags

amirzohrevand wrote:
Bunuel wrote:
hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

I make the answer C.

7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.

Hi Bunuel
I understand your logic, but what if we count the numbers without 7 such that we have a 3 digit number that can be made by numbers {0,1,2,3,4,5,6,8,9} and then subtract the result from 1000. given this logic, 1000-9*9*9= 271

This gives numbers with 7's but some of them will have more than one 7.
_________________
Intern  Joined: 24 Jun 2012
Posts: 10

Show Tags

2
1
My method.

7 can be in units tens or hundres digit.

In units digit - 7 , 17, 27,...87,97 - so 10 times for every 100. so 10*10 for 1000 numbers = 100

In tens digit - 70,71,72,73.....79 - so 10 times for every 100. so 10*10 for 1000 numbers = 100

In hundreds digit - 700,701,702,703....799 - so there are - 100

no other place where 7 can appear.
total = 100+100+100 = 300
Math Expert V
Joined: 02 Sep 2009
Posts: 58318

Show Tags

1
mystseen 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 first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. 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 $$200+100=300$$.

Bunuel - aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271

No, I'm not double counting and the answer is correct. 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.
_________________
Math Expert V
Joined: 02 Aug 2009
Posts: 7953

Show Tags

Bunuel wrote:
mystseen 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 first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. 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 $$200+100=300$$.

Bunuel - aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271

No, I'm not double counting and the answer is correct. 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.

Hi Bunuel,
you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person..
7 at tens place=70,71,72...79=10 places
_________________
CEO  S
Joined: 20 Mar 2014
Posts: 2603
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)

Show Tags

chetan2u wrote:

Hi Bunuel,
you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person..
7 at tens place=70,71,72...79=10 places

Its merely a typo. Bunuel 's solution does capture 10 values and 71-79 can not be 10 values. It must have included 70 to make it 10 values with 7 at the ten's place.
Math Expert V
Joined: 02 Aug 2009
Posts: 7953

Show Tags

1
Engr2012 wrote:
chetan2u wrote:

Hi Bunuel,
you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person..
7 at tens place=70,71,72...79=10 places

Its merely a typo. Bunuel 's solution does capture 10 values and 71-79 can not be 10 values. It must have included 70 to make it 10 values with 7 at the ten's place.

Hi buddy,
few points for you..
1) i have written it is 'missed out' and not a 'mistake'...
2) counting 7 in ten places is a very average thing to know and i believe if you and i can see that then bunuel must have done a 'typo' error and that is why i have mentioned 'missing out'.
3) for any one counting 7 in tens place from 71 to 79 can count ten taking two 7s in 77 and that is what someone has written earlier.
4) the essence of comment was to include 70 in the list..
Regards
_________________
Current Student B
Joined: 23 Nov 2016
Posts: 70
Location: United States (MN)
GMAT 1: 760 Q50 V42 GPA: 3.51

Show Tags

1
If you got 271 (like me), you were most likely tricked .

This question: how many times does 7 appear?

"Typical" question of this type: how many digits with a 7 appear?

The typical question doesn't allow you to count 777 three times (or 77 twice); this question does.
Current Student B
Joined: 23 Nov 2016
Posts: 70
Location: United States (MN)
GMAT 1: 760 Q50 V42 GPA: 3.51

Show Tags

Saurav Arora wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Consider numbers from 1 to 1000 written as follows:

1. 0001

2. 0002

3. 0003

...

1000. 1000

We still have 1000 numbers. However, we used 4 digits per number, hence used total of 4∗1000=4000 digits.
4000/10=400 times.

What's wrong with the above inference?
Thanks.

It doesn't work because you have 4 digits (and aren't accounting for 1001 to 9999). The concept applies as follows:

1 digit (0 to 9): 10 numbers, 1 digit per number = 10 digits --> 10/(10 digits) = each digit used 1 time
2 digits (00 to 99): 10*10=100 numbers, 2 digits per number = 200 digits --> 200/(10 digits) = each digit used 20 times
3 digits (000 to 999): 10*10*10=1000 numbers, 3 digits per number = 3000 digits --> 3000/(10 digits) = each digit used 300 times
4 digits (0000 to 9999): 10*10*10*10=10000 numbers, 4 digits per number = 40000 digits --> 40000/(10 digits) = each digit used 4000 times
..
n digits: 10^n numbers, n digits per number = n*10^n digits --> (n*10^n)/10 = each digit used n*10^(n-1) times
Intern  B
Joined: 23 Jan 2017
Posts: 21

Show Tags

1
POSSIBILE COMBO NUMBER OF 7's use X as every single digit number except 7 -->X=0,1,2,3,4,5,6,8,9 --> 9 POSSIBLE NUMBERS
7XX 81
XX7 81
X7X 81
X77 9x2
77X 9x2
7X7 9x2
777 3
_____
300
Intern  B
Joined: 08 Jul 2016
Posts: 36
Location: Singapore
GMAT 1: 570 Q43 V25 GMAT 2: 640 Q42 V36 WE: Underwriter (Insurance)

Show Tags

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "We have 1000 numbers. We used 3 digits per number, hence used total of 3∗1000=30003∗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) 300010=300300010=300 times."

how have we used 3 digits per number? there are single digit numbers, 2 digits and 3 digit numbers? please explain.
Math Expert V
Joined: 02 Sep 2009
Posts: 58318

Show Tags

prachigautam wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "We have 1000 numbers. We used 3 digits per number, hence used total of 3∗1000=30003∗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) 300010=300300010=300 times."

how have we used 3 digits per number? there are single digit numbers, 2 digits and 3 digit numbers? please explain.

For this approach we are writing single or two-digit numbers using three digits. For example, 3 is written as 003 and 25 is written as 025. This is shown in the solution. If this solution is not clear, you can refer to two other solution given there OR read solution provided by others in this thread OR check other discussion of this question here: https://gmatclub.com/forum/how-many-tim ... 99914.html
_________________
Intern  B
Joined: 13 Sep 2016
Posts: 22
Location: India
GMAT 1: 640 Q48 V29 GPA: 3.5

Show Tags

Bunuel wrote:
mystseen 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 first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. 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 $$200+100=300$$.

Bunuel - aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271

No, I'm not double counting and the answer is correct. 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.

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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 58318

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.
_________________ Re: M01-10   [#permalink] 24 Dec 2017, 01:37

Go to page    1   2    Next  [ 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  