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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?

started using the same method, but got stuck in the middle of the road.

Bunuel, could you please clarify if slot method can be applied to this problem. Thanks in advance.

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?

started using the same method, but got stuck in the middle of the road.

Bunuel, could you please clarify if slot method can be applied to this problem. Thanks in advance.

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.

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
_________________

......................................................................... +1 Kudos please, if you like my post

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

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18 Jul 2014, 14:23

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vad3tha wrote:

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

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

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12 Dec 2014, 00:30

Bunuel wrote:

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.

HI Bunuel,

I also applied same approach. I got 19 7's per 100. here you have mentioned 20 so why are we counting 77 2 times?

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

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28 Mar 2015, 00:39

Another approach: Think of the digits in the form: abcd where a can only be 0 (since 1000 doesn't contain any 7s) and b,c, and d can be any digit from 0-9.

Written Once Ones Place=(1*9*9*1)*1=81 Tens Place= (1*9*1*9)*1=81 Hundreds Place= (1*1*9*9)*1=81

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

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31 Mar 2015, 00:26

Let's take numbers from 700 till 799: We have 100 numbers containing 7 as a hundredth digit We have 10 additional 7 if we take the second digit, and 10 more for the units digit Therefore we have 100+10+10=120 7s.

Let's take numbers from 800 till 899: We have 10 numbers with 7 if we take the second digit, and 10 more for the units digit for a total of 20. The same can be applied for the rest of the hundredth which are 8: 000, 100, 200, 300, 400, 500, 600, 900 So in total we have 9*20+120=300

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

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03 Jul 2015, 09:34

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.

I get how the answer is 300 and that you intentionally start with 000 instead of 001... but I disagree that no digit has preference over another (since the number range is 1-1000 and not 000-999, so the former's number of digits matters the most, esp. if they ask how many 1s digits there are). The 1 digit is used 301 times, not 300 (you have to include the 1 in 1,000). The 0 digit does show up 300 times, but it assumes that we use that method. Even if we assume the method, it depends on which segment of the numbers you look at, esp. if the GMAT asks a variation of this question that shows this difference as outlined below (in which case, you should start with 001 and not 000 so there are 3 less 0s in the first 99 numbers, compared with the three extra 0s in the very last number). So, for the 0s, you can't simply say that 001-099 has the same number of 0s as the 1s in the range 100-199. That gives us a total of 3001 digits.

Yet, even this total is misleading... if the GMAT asks how many times the number 0 shows up in any of the digits from 1-1000, you wouldn't start with "001". Rather, you'd start with "1". Then, the number of 0's in total is 192, and consequently, the total number of digits is 2893. Only the digits 2-9 show up the same number of times between 1-1000. If you want to be a bit more conservative about it, you should only use the method when the same number of digits are used for the lower and upper limits of the range (e.g. 10-99; 100-999), then use the method again for the single digits, the tens digits, and the thousand digits, and add it all up.

The "1" digit shows up this many times per given interval below:

001 - 099 x20 100 - 199 x120 = 100, (looking at just the units/tens digits) 01, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91, all the 1s in the hundred digit is 100-1s 200 - 299 x20 300 - 399 x20 400 - 499 x20 500 - 599 x20 600 - 699 x20 700 - 799 x20 800 - 899 x20 900 - 999 x20 1000 x1

Total = 301

The "0" digit shows up this many times per given interval below, though you should not use this method for the aforementioned reason:

001 - 099 x117 = (looking at just the units/tens digits) 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40, 50, 60, 70, 80, 90, all the 0s in the hundred digit is 99-0s (it's 99 since we can't include 000 and there are only 99 numbers - if we did include 000 it would be 100-0s in the hundred digit + the two other 0s in the number 1000) 100 - 199 x20 200 - 299 x20 300 - 399 x20 400 - 499 x20 500 - 599 x20 600 - 699 x20 700 - 799 x20 800 - 899 x20 900 - 999 x20 1000 x3

Total = 300

Conversely, if they ask how many times the number 0 shows up in each digit from 1-1000, it would be:

\([\frac{(10^n)}{10}]*n - 111*(1 - \frac{x}{x})\), where n is the total number of complete set of digits from 100 onward (so, up to 999 counts as 3, but up to 1,000 doesn't count as 4 as it would need to be 9,999 to count as 4), and x is the number that you want to find repeating (e.g. 0-9). I made up this equation but I think it works... you just need to ignore the undefined part and make it equal to 0 if you were to select 0...

So, if you try it for 1 - 99,999,999, and see how many times the number 4 appears, then it's:

\([\frac{(10^8)}{10}]*8 - 111*(1 - \frac{5}{5}) = 8*10^7\) times the number 4 shows up between 1 and 99,999,999 inclusive.

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

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17 Oct 2015, 04:10

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.

my approach to this question was from permutation view

I took numbers from 0 to 999 (instead of from 1 to 1000), for easiness we are taking here 0 to be as "000" , i.e. all digits as three digits, we can not neglect 0's in the beginning

now we have three places, and every digit from 0 to 9 has equal chance to come on every place so, out of total 1000 numbers 7 comes in hundreds place 100 times out of total 1000 numbers 7 comes in tens place for 100 times and out of total 1000 numbers 7 comes in once place for 100 times

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

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31 Jan 2016, 15:36

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.

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

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26 Nov 2016, 08:25

I think this is easiest to think with the counting formula/boxes.

How many times 7 appears b/w 1-1000. Subtract 1 from 1000 to get 999. Now you are working with a 3 digit number.

Case 1: _ _ _ First digit of the number will be 7. You have 10 digits to pick from but only 1 is 7. (7XX) You have 10 options for the second digit. You have 10 options for the third digit. Case 1 total: 1x10x10 = 100

Case 2: _ _ _ Second digit of the number will be 7. You have 10 digits to pick from but only 1 is 7. (X7X) You have 10 options for the first digit. You have 10 options for the third digit. Case 2 total: 10x1x10 = 100

Case 3: _ _ _ Last digit of the number will be 7. You have 10 digits to pick from but only 1 is 7. (XX7) You have 10 options for the first digit. You have 10 options for the second digit. Case 3 total: 10x10x1 = 100

Add all case totals together: Case 1(100) + Case 2(100) + Case 3(100) = 300 final answer.

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

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01 Jun 2017, 03:07

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

We can solve using permutation (order does matter in numbers):

Possibilities include numbers of 3 digits (from 1 to 999) Either one of the 3 digits has to be 7 (Permutation of a 1 P 3, no repetition) = 3!/(3-1)!=3 The 2 remaining digits can be any other number from 0 to 9 (Permutation of 2 numbers from 10, with repetition = 10^2) Total = 3 x 10^2 = 300

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