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Re M0110 [#permalink]
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15 Sep 2014, 23:15
Official Solution: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 Many approaches are possible. For example: Approach #1: 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) \(\frac{3000}{10}=300\) times. Approach #2: There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately. Onedigit numbers: 7 is the only onedigit number. Twodigit numbers: 7 could be the tens digit or the units digit. Case 1: 7 is the tens digit. There are 10 ways to place 7 as the tens digit of a twodigit number. Case 2: There are 9 ways to place the units digit. Thus, for twodigit numbers we have: \(10+9=19\) numbers that contain a 7. Threedigit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times. Approach #3: 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
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Re: M0110 [#permalink]
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07 Nov 2014, 09:07
Dear experts, Approach #1:What if: Consider numbers from 1 to 1000 written as follows: 1. 0001 2. 0002 3. 0003 ... 1000. 1000 We still have 1000 numbers. However, we used 4 digits per number, hence used total of \(4*1000=4000\) digits. \(\frac{4000}{10}=400\) times. What's wrong with the above inference? Thank you so much! Bunuel wrote: Official Solution:
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
Many approaches are possible. For example: Approach #1: 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) \(\frac{3000}{10}=300\) times. Approach #2: There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately. Onedigit numbers: 7 is the only onedigit number. Twodigit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a twodigit number. Case 2: There are 10 ways to place the second digit, i.e. 09. Remember that we have counted 07 already. Thus, for twodigit numbers we have: \(10+9=19\) numbers that contain a 7. Threedigit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times. Approach #3: 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



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Re: M0110 [#permalink]
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04 Feb 2015, 12:34
Why doesn't the following approach work?
Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 9072 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900648 = 252. Therefore, 7 appears 252 times between 100 and 999.
252 + 19 = 271.
I make the answer C.



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Re: M0110 [#permalink]
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05 Feb 2015, 02:16
hessen923 wrote: Why doesn't the following approach work?
Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 9072 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900648 = 252. Therefore, 7 appears 252 times between 100 and 999.
252 + 19 = 271.
I make the answer C. 7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times. Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.
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Re: M0110 [#permalink]
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13 Apr 2015, 22:26
possibilities are X X 7 = 9*9*1 X 7 X = 9*1*9 7 X X = 1*9*9 7 7 X = 1*1*9 7 X 7 = 1*9*1 X 7 7 = 9*1*1 777 = 1 if we add all of them = 81+81+81+9+9+9+1=271 where did i go wrong??
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WaterFlowsUp wrote: possibilities are
X X 7 = 9*9*1 X 7 X = 9*1*9 7 X X = 1*9*9 7 7 X = 1*1*9 7 X 7 = 1*9*1 X 7 7 = 9*1*1 777 = 1 if we add all of them = 81+81+81+9+9+9+1=271 where did i go wrong?? This is how I did..same approach as you took...this might help you understand what went wrong _ single digit > 1 possibility _ _ two digits > (1 possibility * 10 possibilities) + (9 possibilities * 1 possibility) > (9, as we cannot place a 0) = 19 _ _ _ three digits > (1 possibility * 10 possibilities * 10 possibilities) + (9 possibilities * 1 possibility * 10 possibilities) + (9 possibilities * 10 possibilities * 1 possibility) = 100 + 90 + 90 = 280 add up, 1 + 19 + 280 > 300



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Re: M0110 [#permalink]
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29 Aug 2015, 01:04
Bunuel wrote: hessen923 wrote: Why doesn't the following approach work?
Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 9072 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900648 = 252. Therefore, 7 appears 252 times between 100 and 999.
252 + 19 = 271.
I make the answer C. 7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times. Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's. Hi BunuelI understand your logic, but what if we count the numbers without 7 such that we have a 3 digit number that can be made by numbers {0,1,2,3,4,5,6,8,9} and then subtract the result from 1000. given this logic, 10009*9*9= 271



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Re: M0110 [#permalink]
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29 Aug 2015, 01:25
amirzohrevand wrote: Bunuel wrote: hessen923 wrote: Why doesn't the following approach work?
Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 9072 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900648 = 252. Therefore, 7 appears 252 times between 100 and 999.
252 + 19 = 271.
I make the answer C. 7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times. Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's. Hi BunuelI understand your logic, but what if we count the numbers without 7 such that we have a 3 digit number that can be made by numbers {0,1,2,3,4,5,6,8,9} and then subtract the result from 1000. given this logic, 10009*9*9= 271 This gives numbers with 7's but some of them will have more than one 7.
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Re: M0110 [#permalink]
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30 Aug 2015, 00:47
*** three digits
lets keep right most as 7 this time two can be be filled any 10 digits that includes 0 and 7 also, because we can have to take from 1 to 1000 e.g. 007, 107, 117... so on where 7 at right most.... 10*10 = 100 similarly keep 7 at middle place we more 100 now keep 7 and left most we 100 100+100+100 = 300



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Re: M0110 [#permalink]
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19 Sep 2015, 10:43
My method.
7 can be in units tens or hundres digit.
In units digit  7 , 17, 27,...87,97  so 10 times for every 100. so 10*10 for 1000 numbers = 100
In tens digit  70,71,72,73.....79  so 10 times for every 100. so 10*10 for 1000 numbers = 100
In hundreds digit  700,701,702,703....799  so there are  100
no other place where 7 can appear. total = 100+100+100 = 300



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Re: M0110 [#permalink]
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20 Sep 2015, 08:19
So I made the same mistake and counted the number of integers with a 7 (271) but should have counted the number of times a 7 occurs. I understand approach 2 and 3, but not approach #1. If I count the number of digits from 11000 inclusive this would be:
single digit numbers: 1, 2, ...9 gives 9 numbers with 1 digit > 9x1=9 digits twodigit numbers: 10, 11, ... 99 gives 90 numbers with 2 digits > 90x2=180 digits threedigit numbers: 100, 101, ...999 gives 900 numbers with 3 digits > 900x3=2700 digits fourdigit numbers: 1000. Only a single number with 4 digits> 4 digits
This sums to 9+180+2700+4=2893 digits and not 3000. What am I missing?



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Re: M0110 [#permalink]
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Re: M0110 [#permalink]
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09 Oct 2015, 09:53
Bunuel wrote: Official Solution:
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
Many approaches are possible. For example: Approach #1: 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) \(\frac{3000}{10}=300\) times. Approach #2: There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately. Onedigit numbers: 7 is the only onedigit number. Twodigit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a twodigit number. Case 2: There are 10 ways to place the second digit, i.e. 09. Remember that we have counted 07 already. Thus, for twodigit numbers we have: \(10+9=19\) numbers that contain a 7. Threedigit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times. Approach #3: 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 Bunuel  aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271



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Re: M0110 [#permalink]
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11 Oct 2015, 05:11
mystseen wrote: Bunuel wrote: Official Solution:
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
Many approaches are possible. For example: Approach #1: 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) \(\frac{3000}{10}=300\) times. Approach #2: There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately. Onedigit numbers: 7 is the only onedigit number. Twodigit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a twodigit number. Case 2: There are 10 ways to place the second digit, i.e. 09. Remember that we have counted 07 already. Thus, for twodigit numbers we have: \(10+9=19\) numbers that contain a 7. Threedigit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times. Approach #3: 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 Bunuel  aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271 No, I'm not double counting and the answer is correct. The question asks " How many times will the digit 7 be written when listing the integers from 1 to 1000?" not how many number will have 7.
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Re: M0110 [#permalink]
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25 Nov 2015, 22:29
Bunuel wrote: mystseen wrote: Bunuel wrote: Official Solution:
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
Many approaches are possible. For example: Approach #1: 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) \(\frac{3000}{10}=300\) times. Approach #2: There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately. Onedigit numbers: 7 is the only onedigit number. Twodigit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a twodigit number. Case 2: There are 10 ways to place the second digit, i.e. 09. Remember that we have counted 07 already. Thus, for twodigit numbers we have: \(10+9=19\) numbers that contain a 7. Threedigit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times. Approach #3: 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 Bunuel  aren't you double counting 77 in approach #3? Answer should be 19 * 9 + 100 = 271 No, I'm not double counting and the answer is correct. The question asks " How many times will the digit 7 be written when listing the integers from 1 to 1000?" not how many number will have 7. Hi Bunuel, you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person.. 7 at tens place=70,71,72...79=10 places
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Re: M0110 [#permalink]
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26 Nov 2015, 10:48
chetan2u wrote: Hi Bunuel, you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person.. 7 at tens place=70,71,72...79=10 places Its merely a typo. Bunuel 's solution does capture 10 values and 7179 can not be 10 values. It must have included 70 to make it 10 values with 7 at the ten's place.



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Re: M0110 [#permalink]
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26 Nov 2015, 20:46
Engr2012 wrote: chetan2u wrote: Hi Bunuel, you have missed counting 70 as a number having digit 7 in tens place and hence the doubt of the person.. 7 at tens place=70,71,72...79=10 places Its merely a typo. Bunuel 's solution does capture 10 values and 7179 can not be 10 values. It must have included 70 to make it 10 values with 7 at the ten's place. Hi buddy, few points for you.. 1) i have written it is 'missed out' and not a 'mistake'... 2) counting 7 in ten places is a very average thing to know and i believe if you and i can see that then bunuel must have done a 'typo' error and that is why i have mentioned 'missing out'. 3) for any one counting 7 in tens place from 71 to 79 can count ten taking two 7s in 77 and that is what someone has written earlier. 4) the essence of comment was to include 70 in the list.. Regards
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Re: M0110 [#permalink]
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28 Apr 2016, 21:40
Would this be a possible way of solving this problem?
XX7 We can have 10 various digits = 09 instead of X
And there are 3 possible positions for 7: {7XX, X7X, XX7}
P(2,3)*C(1,10)*C(1,10)
Hence 3*10*10=300.



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Re: M0110 [#permalink]
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22 May 2016, 07:58
Dear Bunuel, I think there is a typo in the explanation of approach 2 below: I think there are 10 numbers meetign the first case which is " Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a twodigit number. " Those numbers are : 70, 71, 72, 73, 74, 75, 76, 77, 78, 79. So the total number of 2 digit numbers become: 10 + 9 = 19. Regards, Ankush Bunuel wrote: Official Solution:
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
Many approaches are possible. For example: Approach #1: 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) \(\frac{3000}{10}=300\) times. Approach #2: There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately. Onedigit numbers: 7 is the only onedigit number. Twodigit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a twodigit number.Case 2: There are 10 ways to place the second digit, i.e. 09. Remember that we have counted 07 already. Thus, for twodigit numbers we have: \(10+9=19\) numbers that contain a 7. Threedigit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times. Approach #3: 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







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