Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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

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

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.

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

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

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.

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.

I think this is a high-quality question and I don't agree with the explanation.

Hi amar.igtr,

Thank you for posting. One request though: please be specific, why don't you agree with the solution? So, far you've posted two similar replies and none of them were correct. Thank you.
_________________

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.

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 _________________