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Re: M0110
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24 Dec 2017, 01:37
Siddhuftr wrote: ones  x  1 time tens  x7 + 7x  9*1 + 1*10 = 19 times Hundreds  xx7 + x7x + 7xx = 9*10*1 + 9*1*10 + 1*10*10 = 190 Total = 300
I am unable to understand , how double counting of 77 , 777 is avoided ? 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. So, for example, for number 777, it should be counted as three 7's not one.
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Re: M0110
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21 May 2018, 09:25
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.
Answer: D Hello BunuelCould you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000? I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused. Thank you,



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Re: M0110
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21 May 2018, 21:28
vitorcbarbieri 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.
Answer: D Hello BunuelCould you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000? I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused. Thank you, Since neither 0 not 1000 has 7 in it, then excluding/including these values into the range won't change the answer.
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Re: M0110
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03 May 2019, 22:15
Bunuel chetan2uThe answer will be 300 for all the digits except 0 and 1, right? For 0, the answer will be 303 (Considering 3 zeroes in 1000) and for 1, the answer will be 301 (Considering one 1 in 1000). Let me know if my understanding is correct?
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Re: M0110
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29 May 2019, 16:25
Hey Bunuel, what is meant by "In 10 hundreds 7" in approach 3? I'm having a hard time understanding this.
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Re: M0110
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11 Sep 2019, 11:09
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 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 Hi Bunuel, In approach # 1, i didn't quite understand why did we use 3 digit numbers. Can you please explain? Apologies if it's a silly question.



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17 Sep 2019, 05:54
swatjazz 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 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 Hi Bunuel, In approach # 1, i didn't quite understand why did we use 3 digit numbers. Can you please explain? Apologies if it's a silly question. Not a silly question at all. Let me try to elaborate: this approach worked because when we write the numbers from 0 to 999 in the form XXX each digit take the values from 0 to 9 which provides that in the end all digits are used equal # of times. For the range 100 to 999 it won't be so. We can solve for this range in the following way: XX7  7 in the units place  first digit can take 9 values (from 1 to 9) and second digit can take 10 values (from 0 to 9) > total numbers with 7 in the units place: 9*10=90; X7X  7 in the tens place  first digit can take 9 values (from 1 to 9) and third digit can take 10 values (from 0 to 9) > total numbers with 7 in the tens place: 9*10=90; 7XX  7 in the hundreds place  second digit can take 10 values (from 0 to 9) and third digit can take 10 values (from 0 to 9) > total numbers with 7 in the hundreds place: 10*10=100. TOTAL: 90+90+100=280. Hope it helps.
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