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

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

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

Re: How many times will the digit 7 be written? [#permalink]
17 Jul 2014, 21:23

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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]
28 Aug 2010, 02:22

1

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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! _________________

Re: How many times will the digit 7 be written? [#permalink]
28 Aug 2010, 12:46

1

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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?

============================== Kudos me if you like this method or post! _________________

Re: How many times will the digit 7 be written? [#permalink]
29 Aug 2010, 00:11

1

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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.

Re: How many times will the digit 7 be written? [#permalink]
14 Sep 2010, 14:51

1

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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

Re: How many times will the digit 7 be written? [#permalink]
30 Sep 2010, 14:49

1

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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)

Re: How many times will the digit 7 be written? [#permalink]
11 Jan 2011, 13:27

1

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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. _________________

Re: How many times will the digit 7 be written? [#permalink]
19 Jan 2011, 02:03

1

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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 =

Re: How many times will the digit 7 be written? [#permalink]
14 Jul 2014, 16:54

1

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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]
18 Jul 2014, 13:23

1

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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

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?

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