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Intern
Joined: 02 Oct 2014
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Re: M01-10
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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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Intern
Joined: 27 Feb 2015
Posts: 48
Concentration: General Management, Economics
WE: Engineering (Transportation)
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Re: M01-10
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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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Manager
Joined: 23 Apr 2014
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Location: United States
GPA: 2.75
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Re M01-10
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14 Jul 2016, 11:44
I think this is a high-quality question and I agree with explanation. Third approach was easiest for me to understand.
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Intern
Joined: 14 Jun 2015
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Re M01-10
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31 Jul 2016, 02:33
I think this is a high-quality 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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Manager
Joined: 03 Apr 2015
Posts: 185
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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 (0-9) 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 (0-9) 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 (0-9) 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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Intern
Joined: 06 Aug 2016
Posts: 1
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Re: M01-10
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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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Manager
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GPA: 3.51
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Re: M01-10
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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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Manager
Joined: 23 Nov 2016
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Re: M01-10
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27 Dec 2016, 08:46
Saurav Arora wrote: I think this is a high-quality 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^(n-1) times
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Intern
Joined: 23 Jan 2017
Posts: 21
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Re: M01-10
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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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Intern
Joined: 12 Mar 2016
Posts: 5
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Re: M01-10
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01 May 2017, 18:57
Hi,
Are we not counting 77 twice in Approach 3?
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Manager
Joined: 14 May 2015
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Re: M01-10
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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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Intern
Joined: 22 Mar 2017
Posts: 7
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Re: M01-10
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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 0-9)
7 at tens place, that means total combination will be:- 9*1*10 (first digit can be any number from 1-9, second will only be 7 and third will be any number from 0-9)
7 at unit place, that means total combination will be:- 9*10*1 (first digit can be any number from 1-9, second will any number from 0-9 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 0-9)
7 at unit place total combination will be:- 9*1 (first digit can be any number from 1-9 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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Intern
Joined: 08 Apr 2016
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Re: M01-10
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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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Math Expert
Joined: 02 Sep 2009
Posts: 50730
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Re: M01-10
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14 Jul 2017, 05:48
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Intern
Joined: 24 Aug 2017
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Re: M01-10
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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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Joined: 09 Dec 2015
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Re M01-10
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02 Oct 2017, 20:05
I think this is a high-quality question and I don't agree with the explanation.
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Math Expert
Joined: 02 Sep 2009
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Re: M01-10
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02 Oct 2017, 20:12
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Intern
Joined: 08 Jul 2016
Posts: 38
Location: Singapore
GMAT 1: 570 Q43 V25 GMAT 2: 640 Q42 V36
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Re M01-10
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04 Dec 2017, 00:32
I think this is a high-quality 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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Math Expert
Joined: 02 Sep 2009
Posts: 50730
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prachigautam wrote: I think this is a high-quality 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 two-digit 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/how-many-tim ... 99914.html
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Intern
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Re: M01-10
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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.
One-digit numbers: 7 is the only one-digit number.
Two-digit 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 two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: \(10+9=19\) numbers that contain a 7.
Three-digit 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 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).
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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