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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]

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27 Aug 2010, 14:38
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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
[Reveal] Spoiler: OA

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Re: How many times will the digit 7 be written? [#permalink]

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27 Aug 2010, 14:51
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Expert's post
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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.

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Re: How many times will the digit 7 be written? [#permalink]

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28 Aug 2010, 14:18
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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]

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14 Sep 2010, 06:56
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Expert's post
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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.

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.

Hope it helps.
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02 Sep 2010, 13:12
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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]

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10 Oct 2012, 05:56
3
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MOST EFFICIENT METHOD USING COMBINATORIX

As Bunuel said, think of each number as XXX ie 001 to 999

Numbers with one 7.

Either 7XX, X7X or XX7, so 3 x 1C9 x 1C9 = 3x9x9 = 243

Numbers with two 7's (nb question asks how many sevens are displayed, so this answer must be doubled)

Either 77X, 7X7 or X77, so 3x 1C9 = 27 (or 54 once you double it)

Numbers with three 7's

Only 1 (777). Triple it as there are three 7,s = 3

Add all together: 243+54+3 = 300

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Re: How many times will the digit 7 be written? [#permalink]

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28 Aug 2010, 13:46
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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]

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11 Jan 2011, 14:27
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Expert's post
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]

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17 Jul 2014, 22:23
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The first time I did the question, my answer was C : 271 since I thought the question is "how many numbers from 0 - 1000, that contain the 7 digit". So I counted 77, 707,717,....,777,...797, 771, 772, 773,..., or 779 as one.
That may be the reason why 23% of people who answered the question decided to pick answer C.
That is really interesting question
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Re: How many times will the digit 7 be written? [#permalink]

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18 Jul 2014, 01:08
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It is very interesting question, thanks for the all the suggestion but Banuel is relly a rock-star..
how to promote an app

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Re: How many times will the digit 7 be written? [#permalink]

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28 Aug 2010, 03:22
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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]

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29 Aug 2010, 01:11
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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.

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

Thanks!
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Re: How many times will the digit 7 be written? [#permalink]

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14 Sep 2010, 15:51
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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]

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30 Sep 2010, 15:49
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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]

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19 Jan 2011, 03:03
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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]

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13 Sep 2012, 00:21
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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]

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14 Jul 2014, 17:54
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Alternate solution:

Using probability of occurrence = desired outcomes / total outcomes, desired outcomes = probability x total outcomes.

total outcomes = 1000.

probability of occurrence - to simplify this, the question statement could be interpreted as "what is the probability that the digit 7 is picked at least once when 3 digits are chosen at random?"

probability of at least one 7 = 7 in first digit OR 7 in second digit OR 7 in third digit = 1/10 + 1/10 + 1/10 = 3/10

desired outcomes = probability x total outcomes = 3/10 x 1000 = 300.

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Re: How many times will the digit 7 be written? [#permalink]

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18 Jul 2014, 14:23
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The first time I did the question, my answer was C : 271 since I thought the question is "how many numbers from 0 - 1000, that contain the 7 digit". So I counted 77, 707,717,....,777,...797, 771, 772, 773,..., or 779 as one.
That may be the reason why 23% of people who answered the question decided to pick answer C.
That is really interesting question

Yes, I made the same mistake.

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Re: How many times will the digit 7 be written? [#permalink]

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28 Aug 2010, 07:25
excellent approach Bunuel. You are indeed a rockstar.

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Re: How many times will the digit 7 be written? [#permalink]

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14 Sep 2010, 04:45
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.

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.

Kudos [?]: 15 [0], given: 17

Re: How many times will the digit 7 be written?   [#permalink] 14 Sep 2010, 04:45

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