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How many times will the digit 7 be written?

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How many times will the digit 7 be written? [#permalink] New post 27 Aug 2010, 13:38
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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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Re: How many times will the digit 7 be written? [#permalink] New post 27 Aug 2010, 13:51
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seekmba wrote:
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

Can someone explain a simpler approach to this problem?


Many approaches are possible. For example:

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) 3000/10=300 times.

Answer: D.
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Re: How many times will the digit 7 be written? [#permalink] New post 28 Aug 2010, 02:22
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Thanks Bunuel. This is logical answer by you. When I had seen such a question, I counted all the numbers!! It was funny :)
Anyhow, thanks for giving reasons for the correct answer.
You are the ROCKSTAR!
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Re: How many times will the digit 7 be written? [#permalink] New post 28 Aug 2010, 06:25
excellent approach Bunuel. You are indeed a rockstar.
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Re: How many times will the digit 7 be written? [#permalink] New post 28 Aug 2010, 12:46
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Wow, I really like Brunel's method.
But I tried to do it in a different way. Not a very intelligent method, but it sure will not take much time.
First do it from 1-100, then look at the bigger picture.

But I still am facing a problem.

In one to hundred,
7
17
27
37
47
57
67
-
87
97
70
:
:
79

Thats 19 * 10 = 190 7s in the tens place and units places in teh first thousand numbers.

Hundreds place:
700
701
:
:
799


Older 190 + hundreds place 100 7s = 290

Where am I missing the 10 others?


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Re: How many times will the digit 7 be written? [#permalink] New post 28 Aug 2010, 13:18
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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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Re: How many times will the digit 7 be written? [#permalink] New post 29 Aug 2010, 00:11
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pradosamant wrote:
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.




Aahhhhhh... Yessss.....
Good... +1 to you! :)

Thanks!
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counting [#permalink] New post 02 Sep 2010, 12:12
How many times will the digit 7 be written when listing the integers from 1 to 1000?

110
111
271
300
304

any easy way to do this question?
here is how i did it...
let xyz be a 3-digit#, first, let Z=7, we have 10 options ( 0-9)for x and 10 options for y, then we get 10 x 10 =100, same logic is applied when y =7 and x = 7, we get 300 times.
is it correct?
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Re: How many times will the digit 7 be written? [#permalink] New post 14 Sep 2010, 03:45
Bunuel wrote:
seekmba wrote:
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

Can someone explain a simpler approach to this problem?


Many approaches are possible. For example:

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) 3000/10=300 times.

Answer: D.



Thanks Bunuel... But please explain how can i proceed using permutation to solve this one for my understanding...I am able to get a count =280 but not 300 using permutation method.
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Re: How many times will the digit 7 be written? [#permalink] New post 14 Sep 2010, 05:56
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utin wrote:
Bunuel wrote:
seekmba wrote:
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

Can someone explain a simpler approach to this problem?


Many approaches are possible. For example:

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) 3000/10=300 times.

Answer: D.



Thanks Bunuel... But please explain how can i proceed using permutation to solve this one for my understanding...I am able to get a count =280 but not 300 using permutation method.


Another approach:

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.

Hope it helps.
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Re: How many times will the digit 7 be written? [#permalink] New post 14 Sep 2010, 14:51
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I made simple thing complex by using permutations here. I got answer but kicked myself after reading simple concept presented by bunuel. I fixed 7 number on specific place and counted how many numbers can be generated with other digits

Single digit = 1
Double digit = 9C1+8C1 + 2 = 19 (7 _ and _7 and 77)
Three digit = 9C1*9C1 + 8C1*9C1 + 8C1*9C1 ( 7 _ _, _ 7 _; _ _ 7)
2( 9C1 + 9C1 + 8C1) (77_; 7_7; _77)
+ 3 (777)
1 + 19 + (81+72+72) + 52 + 3 = 300
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Re: How many times will the digit 7 be written? [#permalink] New post 30 Sep 2010, 14:49
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First note that we only need consider the numbers 0 to 999. (In other words, we
consider 3-digit numbers which may start with a zero; e.g., 11 = 011.) There are
3 possible places for a 7 to appear. We can proceed by counting the number of times
a 7 appears in each place. So fix a place for a 7 to appear; then for the remaining 2
digits, there are 10^2 possibilities. So the total is 3 · 10^2.


To generalize the case, if you see:
"How many times is the digit X written when listing all numbers from 1 to Y"
The total is (#of digits in Y-1)*10^(#of digits in Y-2)
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Re: How many times will the digit 7 be written? [#permalink] New post 11 Jan 2011, 13:27
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nonameee wrote:
Bunuel, can you please explain the logic here? I don't understand it.

Quote:
Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times...


Not sure what can I add to this... It just means that out of 3000 digits we used to write down first 1000 numbers (000, 001, 002, 003, ..., 999), each digit from 0 to 9 is used equal number of times, how else? Thus we used each digit 3000/10=300 times.
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Re: How many times will the digit 7 be written? [#permalink] New post 19 Jan 2011, 01:29
Bunuel, thanks. Now I got it.
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Re: How many times will the digit 7 be written? [#permalink] New post 19 Jan 2011, 02:03
10 times 7 on units place in first one hundred so 100 times together +
10 times 7 on tens place in first one hundred so another 100 times together +
100 times 7 on hundreds place =

300
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Re: How many times will the digit 7 be written? [#permalink] New post 13 Feb 2011, 14:36
Thanks Bunnel. I like your approach. It is best, rather counting, which I did;-)
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Re: How many times will the digit 7 be written? [#permalink] New post 28 Sep 2011, 09:51
Bunuel wrote:
seekmba wrote:
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

Can someone explain a simpler approach to this problem?


Many approaches are possible. For example:

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) 3000/10=300 times.

Answer: D.



Hi Bunnel,
Suppose we have to find no of times 7 is found between 11 and 1347.
How can we find it with a logical explanation as you have given above?

Thanks,
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Re: How many times will the digit 7 be written? [#permalink] New post 28 Sep 2011, 23:22
I used the counting method, but it did not take me too long; nearly 80 secs. I counted the 7s only from 0-99 (20 in all) and then applied it to the subsequent sets, with a careful consideration for the set 700-799.

0-99 : 20
100-199 : 20
200-299 : 20
300-399 : 20
400-499 : 20
500-599 : 20
600-699 : 20
700-799 : 20 + 100
800-899 : 20
900-999 : 20

so, Total = (20 x 10) + 100 = 300

Ans. D

But, Bunuel's solution is undoubtedly the best, quite logical and NOT error prone. If one can think so logically during the test, Bunuel's approach will rock.

I will remember iPinnacle's tip too
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Re: How many times will the digit 7 be written? [#permalink] New post 12 Sep 2012, 23:21
7 does not occur in 1000. So we have to count the number of times it appears between 1
and 999. Any number between 1 and 999 can be expressed in the form of xyz where
0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other
9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second
or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and
3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with
the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or
m second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300
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Re: How many times will the digit 7 be written? [#permalink] New post 20 Sep 2012, 20:10
from 0 - 100 , 7 shows up 20 times ( 07,17,27,37,47,57,67,70,71,72,73,,74,75,76,77,78,79,87,97) therefore from 0-1000 (10 sets of 20 each) we will have 20 x 10 = 200 (the times 7 will show either once or twice ie seven shows up by itself 19 times, and once twice in 77 ) ... Now in the series from 700 - 799 included SEVEN shows up Once in every number and either once or twice in 20 numbers. , therefore we will have 100 additional seven's for that series (700-799) therefore it is 100 + 20 = 120 .

Thus we have (20 x 9) + 120 = 300 ...

( Seven will show up once in 19 numbers between 0 - 100 , but twice in one number ie 77 Therefore the number of times seven appears in the first 100 numbers is 20 (19+1) ... This can be multiplied by 9 (leave the series from 700-799) and we get 180 ... Add 120 to the 180 and we would have included all the times seven shows up ie. in addition to the numbers where seven shows up twice we would have also included those in which seven shows up three times. So the answer is 180+120 = 300 ... )


Answer D ...
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Re: How many times will the digit 7 be written?   [#permalink] 20 Sep 2012, 20:10
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