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M01-10

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M01-10  [#permalink]

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New post 16 Sep 2014, 00:14
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A
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C
D
E

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  95% (hard)

Question Stats:

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

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Re M01-10  [#permalink]

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New post 16 Sep 2014, 00:15
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 \(200+100=300\).


Answer: D
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Re: M01-10  [#permalink]

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New post 04 Feb 2015, 13:34
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.
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Re: M01-10  [#permalink]

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New post 05 Feb 2015, 03:16
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.
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Re: M01-10  [#permalink]

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New post 13 Apr 2015, 23:26
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??
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Re: M01-10  [#permalink]

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New post 12 Aug 2015, 03:42
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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
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Re: M01-10  [#permalink]

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New post 29 Aug 2015, 02:04
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
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Re: M01-10  [#permalink]

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New post 29 Aug 2015, 02:25
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.
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Re: M01-10  [#permalink]

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New post 19 Sep 2015, 11:43
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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
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Re: M01-10  [#permalink]

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New post 11 Oct 2015, 06:11
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 \(200+100=300\).


Answer: D


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.
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Re: M01-10  [#permalink]

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New post 25 Nov 2015, 23:29
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 \(200+100=300\).


Answer: D


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
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Re: M01-10  [#permalink]

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New post 26 Nov 2015, 11:48
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.
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Re: M01-10  [#permalink]

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New post 26 Nov 2015, 21:46
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
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Re: M01-10  [#permalink]

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New post 27 Dec 2016, 09:28
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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.
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Re: M01-10  [#permalink]

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New post 27 Dec 2016, 09:46
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
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Re: M01-10  [#permalink]

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New post 14 Feb 2017, 08:22
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
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New post 04 Dec 2017, 01:32
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.
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New post 04 Dec 2017, 01:52
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
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Re: M01-10  [#permalink]

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New post 24 Dec 2017, 01:31
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 \(200+100=300\).


Answer: D


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 ?
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Re: M01-10  [#permalink]

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New post 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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