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Re: M0110 [#permalink]
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30 Jun 2016, 09:00
Using basic counting principles: _ _ _ Total Numbers to be considered = 000 to 999 Case1: _ _ 7 Digits 0  9 can take first place in 10 ways. Similarly 0  9 digits can fill up the second place in 10 ways. Total ways = m * n = 10 * 10 = 100 Case2: _ 7 _ Total 100 ways Case3: 7 _ _ Total 100 ways Case1+2+3 =



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Re: M0110 [#permalink]
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07 Jul 2016, 08:27
Bunuelhello can we do it this way ? since there are 10 digits and repetitions can be allowed . keeping 7 in hundreds place >> 7 __ __ { 1*10*10} keeping 7 in tens place >> __ 7[u]__ { 10*1*10} keeping 7 in units place >> __ __ [u]7 {10*10*1} adding above 3 we get 300 also 3c1 * 10*10



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Re M0110 [#permalink]
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14 Jul 2016, 11:44
I think this is a highquality question and I agree with explanation. Third approach was easiest for me to understand.



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Re M0110 [#permalink]
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31 Jul 2016, 02:33
I think this is a highquality question and the explanation isn't clear enough, please elaborate. 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. 4000/10=400 times.
What's wrong with the above inference? Thanks.



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is this approach correct there are 3 places x,x,x consider 7 can be placed in each of the position case 1 7 at hundredths place then rest 2 position can be filled by 10 numbers (09) there fore total in 100 ways (1X10X10) (possibilities are in and clause) case2 7 at tenths place then rest 2 position can be filled by 10 numbers (09) there fore total in 100 ways (10X1X10) (possibilities are in and clause) case 3 7 at ones place then rest 2 position can be filled by 10 numbers (09) there fore total in 100 ways (10X10X1) (possibilities are in and clause) as these possibilities are in or clause so adding them up gives 300 ways



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Re: M0110 [#permalink]
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13 Aug 2016, 21:00
Haihai, I believe that's because when you have four digits the numbers could exceed 1000 such as 7,674 and so on. I believe what you did would be the approach to finding out how many times 7 is used between 1 and 10,000.



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Re: M0110 [#permalink]
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27 Dec 2016, 08:28
If you got 271 (like me), you were most likely tricked . This question: how many times does 7 appear? "Typical" question of this type: how many digits with a 7 appear? The typical question doesn't allow you to count 777 three times (or 77 twice); this question does.



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Re: M0110 [#permalink]
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27 Dec 2016, 08:46
Saurav Arora wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. 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. 4000/10=400 times.
What's wrong with the above inference? Thanks. It doesn't work because you have 4 digits (and aren't accounting for 1001 to 9999). The concept applies as follows: 1 digit (0 to 9): 10 numbers, 1 digit per number = 10 digits > 10/(10 digits) = each digit used 1 time 2 digits (00 to 99): 10*10=100 numbers, 2 digits per number = 200 digits > 200/(10 digits) = each digit used 20 times 3 digits (000 to 999): 10*10*10=1000 numbers, 3 digits per number = 3000 digits > 3000/(10 digits) = each digit used 300 times 4 digits (0000 to 9999): 10*10*10*10=10000 numbers, 4 digits per number = 40000 digits > 40000/(10 digits) = each digit used 4000 times .. n digits: 10^n numbers, n digits per number = n*10^n digits > (n*10^n)/10 = each digit used n*10^(n1) times



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Re: M0110 [#permalink]
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14 Feb 2017, 07:22
POSSIBILE COMBO NUMBER OF 7's use X as every single digit number except 7 >X=0,1,2,3,4,5,6,8,9 > 9 POSSIBLE NUMBERS 7XX 81 XX7 81 X7X 81 X77 9x2 77X 9x2 7X7 9x2 777 3 _____ 300



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Re: M0110 [#permalink]
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01 May 2017, 18:57
Hi,
Are we not counting 77 twice in Approach 3?



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Re: M0110 [#permalink]
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06 May 2017, 13:05
(7)=1 9 (7)=9 (7) 10=10 9 10 (7)=90 9 (7) 10=90 (7) 10 10 = 100 total=300 ans:D



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Re: M0110 [#permalink]
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03 Jun 2017, 07:22
For a 3 digit number, 7 can be at the unit place, tens place or hundred place. 7 at hundred place, that means total combination will be: 1*10*10 (first digit can only be seven but second and third can be any from 09) 7 at tens place, that means total combination will be: 9*1*10 (first digit can be any number from 19, second will only be 7 and third will be any number from 09) 7 at unit place, that means total combination will be: 9*10*1 (first digit can be any number from 19, second will any number from 09 and third will be 7) So total 3 digit number with 7 at some place = 100+90+90=280 (You can calculate this for any digit)
For a 2 digit number, 7 can be at unit place, or tens place Going with the same logic as described above: 7 at tens place total combination will be: 1*10 (first digit can only be seven but second can be any from 09) 7 at unit place total combination will be: 9*1 (first digit can be any number from 19 and second digit is 7) So for 2 digit number total combination will be 10+9=19 (You can calculate this for any digit)
For single digit number, there is only one possibility So total is 280+19+1=300. D is the answer.



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Re: M0110 [#permalink]
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14 Jul 2017, 05:45
On counting three digit numbers,total is 280 but this also includes 777 thrice,which means we should subtract 2 777's leading to answer of 298.



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Re: M0110 [#permalink]
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14 Jul 2017, 05:48



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Re: M0110 [#permalink]
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04 Sep 2017, 10:40
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?? in cases 77X, 7X7, X77 and 777 the digit appears 1*1*9 * 2, 1*9*1*2, 9*1*1*2, 1*3. You do this the answer is exactly 300. Note Question is about number of 7, not the numbers which have seven. eg. 777 has 3 sevens. Hope it helps.



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Re M0110 [#permalink]
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02 Oct 2017, 20:05
I think this is a highquality question and I don't agree with the explanation.



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Re: M0110 [#permalink]
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02 Oct 2017, 20:12



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Re M0110 [#permalink]
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04 Dec 2017, 00:32
I think this is a highquality question and the explanation isn't clear enough, please elaborate. "We have 1000 numbers. We used 3 digits per number, hence used total of 3∗1000=30003∗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) 300010=300300010=300 times."
how have we used 3 digits per number? there are single digit numbers, 2 digits and 3 digit numbers? please explain.



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prachigautam wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. "We have 1000 numbers. We used 3 digits per number, hence used total of 3∗1000=30003∗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) 300010=300300010=300 times."
how have we used 3 digits per number? there are single digit numbers, 2 digits and 3 digit numbers? please explain. For this approach we are writing single or twodigit numbers using three digits. For example, 3 is written as 003 and 25 is written as 025. This is shown in the solution. If this solution is not clear, you can refer to two other solution given there OR read solution provided by others in this thread OR check other discussion of this question here: https://gmatclub.com/forum/howmanytim ... 99914.html
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Re: M0110 [#permalink]
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24 Dec 2017, 00:31
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. 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 ?







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