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How many times will the digit 7 be written when listing the integers
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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 M0110
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Re: How many times will the digit 7 be written when listing the integers
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27 Aug 2010, 14:51
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 when listing the integers
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28 Aug 2010, 14:18
in case of counting from 0 to 100 you are missing 77the 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 when listing the integers
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14 Sep 2010, 06:56
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 0100: 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 when listing the integers
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14 Sep 2010, 15:51
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 when listing the integers
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02 Sep 2010, 13:12
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 3digit#, first, let Z=7, we have 10 options ( 09)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 when listing the integers
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30 Sep 2010, 15:49
First note that we only need consider the numbers 0 to 999. (In other words, we consider 3digit 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 Y1)*10^(#of digits in Y2)



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Re: How many times will the digit 7 be written when listing the integers
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10 Oct 2012, 05:56
MOST EFFICIENT METHOD USING COMBINATORIXAs 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 Answer => D
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Re: How many times will the digit 7 be written when listing the integers
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28 Aug 2010, 13:46
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 1100, 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!
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Re: How many times will the digit 7 be written when listing the integers
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11 Jan 2011, 14:27
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 when listing the integers
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13 Sep 2012, 00:21
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 (1digit, 2digits 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 when listing the integers
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28 Aug 2010, 03:22
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 when listing the integers
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14 Jul 2014, 17:54
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 when listing the integers
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17 Jul 2014, 22:23
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 when listing the integers
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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 when listing the integers
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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. 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: the principle of coounting?
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05 Jul 2011, 03:36
Another approach: 7 in unit's place: In every 10 consecutive numbers, 7 will appear once. Since we have 1000 numbers, we have 100 sequences of 10 consecutive numbers (110, 1120 etc) so 7 will appear in unit's place 100 times. 7 in ten's place: In every 100 consecutive numbers, 7 appears in ten's place 10 times (from 70 to 79). We have 10 sequences of 100 consecutive numbers (1100, 101200 etc) so we get that 7 will appear in ten's place 10*10 = 100 times. In every 1000 numbers, 7 will appear in hundred's place 100 times (from 700 to 799). Total = 100 + 100 + 100 = 300
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Re: How many times will the digit 7 be written when listing the integers f
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03 Aug 2011, 00:23
My method is not as simple, but I think it is easier to understand than the previous methods.
First off for single and double digit numbers, lets assume everyone can figure out that there are 20 7's.
Once we get into the hundreds, we know that the 700s will be tricky, but for all the other hundreds, we should be seeing 20 7's in each set of hundred. Other than the 700s, there are only 8 other sets of hundreds (100s, 200s, 300s, etc.). So we multiply 20 * 8 = 160...bringing the total to 180.
So far we have accounted for all the 7's for single and double digit numbers and all hundreds not including the 700s.
For the 700s, we already know that there is at least one 7 from the numbers 700799. So there's 100 right there. At this point the only remaining numbers with 7 are the same ones we identified for single and double digit numbers. (20 of em)
So in the end we have... 20+160+100+20 = 300
compared to the above methods, this is a very crude way of doing this problem



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Re: How many times will the digit 7 be written when listing the integers
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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? Please clarify. Thanks



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Re: How many times will the digit 7 be written when listing the integers
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12 Dec 2014, 06:48
PathFinder007 wrote: 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?
Please clarify.
Thanks The questions asks: how many times will the digit 7 be written ...? Now, how many times is 7 written in 77?
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Re: How many times will the digit 7 be written when listing the integers
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