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:

How many different 7 digit members are their sum of whose digits is even ?

Here is how i approached it.

Sum of the 7 digits is even if the number has 1 or 3 or 5 or 7 even digits There are 5 even digits available 0,2,4,6,8 and 5 odd digits available 1,3,5,7,9. However the first digit in the 7 digit number should not be 0. Otherwise it will become a 6 digit number.

7E 0O = \(5^7-5^6\) There are 5 even numbers available. So each of 7 the digits can be filled with an even number in \(5^7\) ways, but this also contains the numbers having 0 in the first digit location. If first digit has a 0 , rest 6 digits can be filled with an even number in \(5^6\) ways

5E 2O = \(C^7_5*5^5*5^2-C^6_4*5^4*5^2\)

We can first pick the 5 spots having even numbers in \(C^7_5\) ways ,fill them each with even number in \(5^5\) ways and fill the remaining 2 digits with odd numbers in \(5^2\) ways. Again we need to subtract the numbers having 0 in the first digit location If first digit has a 0 , we can pick the 4 spots out of the remaining 6 having even numbers in \(C^6_4\) ways ,fill them each with even number in \(5^4\) ways and fill the remaining 2 digits with odd numbers in \(5^2\) ways.

3E 4O = \(C^7_3*5^3*5^4-C^6_2*5^2*5^4\)

We can first pick the 3 spots having even numbers in \(C^7_3\) ways ,fill them each with even number in \(5^3\) ways and fill the remaining 4 digits with odd numbers in \(5^4\) ways. Again we need to subtract the numbers having 0 in the first digit location If first digit has a 0 , we can pick the 2 spots out of the remaining 6 having even numbers in \(C^6_2\) ways ,fill them each with even number in 5^2 ways and fill the remaining 4 digits with odd numbers in \(5^4\) ways.

1E 6O = \(C^7_1*5^1*5^6-5^6\)

We can first pick the 1 spot having an even number in \(C^7_1\) ways ,fill it with an even number in \(5^1\) ways and fill the remaining 6 digits with odd numbers in \(5^6\) ways. Again we need to subtract the numbers having 0 in the first digit location If first digit has a 0 , we can fill the remaining 6 digits with odd numbers in \(5^6\) ways.

Total no of 7 digit numbers whose sum of digits is even= \(5^7-5^6\)+\(C^7_5*5^7-C^6_4*5^6\)+ \(C^7_3*5^7-C^6_2*5^6\)+\(C^7_1*5^7-5^6\) =\(4*5^6+90*5^6+160*5^6+34*5^6\) =\(288*5^6\)

Zareentaj - What's the OA ? _________________

___________________________________ Please give me kudos if you like my post

How many different 7-digit numbers are there the sum of whose digits are even?

From the first glance I'd say exactly half of the number will have the sum of their digits odd and half even, why should there be more even than odd?

Consider first ten 7-digit integers: 1,000,000-1,000,009 - half has even sum and half odd; Next ten 7-digit integers: 1,000,010-1,000,019 - also half has even sum and half odd; And so on.

There are \(9*10^6\) 7-digit integers, half of it, \(45*10^5\) will have odd sum of their digits, or as crack700 wrote \(288*5^6\), (\(288*5^6=2^5*9*5^6=(9*5)*(2^5*5^5)=45*10^5\)).

I like Bunuel's approach, or you can think of things as follows. Let's start by ignoring the units digit:

* we have 9 choices for the first digit, and 10 choices for each remaining digit, so there are 9*10^5 choices for the first six digits.

* now the sum of the first six digits is either even or odd. If it's even, we need our last digit to be even, so we have 5 choices. If it's odd, we need our last digit to be odd, so we have 5 choices. So no matter how we choose our first six digits, we have 5 choices for our units digit, and the answer must then be 9 * 10^5 * 5 = 45* 10^5. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Bunuel/Ian Can this approach be extended in general i.e., How many N digit integers are there whose sum is even/odd? Answer = 9 x 10^(N-1)/2? _________________

How many different 7 digit members are their sum of whose digits is even ?

Here is how i approached it.

Sum of the 7 digits is even if the number has 1 or 3 or 5 or 7 even digits There are 5 even digits available 0,2,4,6,8 and 5 odd digits available 1,3,5,7,9. However the first digit in the 7 digit number should not be 0. Otherwise it will become a 6 digit number.

7E 0O = \(5^7-5^6\) There are 5 even numbers available. So each of 7 the digits can be filled with an even number in \(5^7\) ways, but this also contains the numbers having 0 in the first digit location. If first digit has a 0 , rest 6 digits can be filled with an even number in \(5^6\) ways

5E 2O = \(C^7_5*5^5*5^2-C^6_4*5^4*5^2\)

We can first pick the 5 spots having even numbers in \(C^7_5\) ways ,fill them each with even number in \(5^5\) ways and fill the remaining 2 digits with odd numbers in \(5^2\) ways. Again we need to subtract the numbers having 0 in the first digit location If first digit has a 0 , we can pick the 4 spots out of the remaining 6 having even numbers in \(C^6_4\) ways ,fill them each with even number in \(5^4\) ways and fill the remaining 2 digits with odd numbers in \(5^2\) ways.

3E 4O = \(C^7_3*5^3*5^4-C^6_2*5^2*5^4\)

We can first pick the 3 spots having even numbers in \(C^7_3\) ways ,fill them each with even number in \(5^3\) ways and fill the remaining 4 digits with odd numbers in \(5^4\) ways. Again we need to subtract the numbers having 0 in the first digit location If first digit has a 0 , we can pick the 2 spots out of the remaining 6 having even numbers in \(C^6_2\) ways ,fill them each with even number in 5^2 ways and fill the remaining 4 digits with odd numbers in \(5^4\) ways.

1E 6O = \(C^7_1*5^1*5^6-5^6\)

We can first pick the 1 spot having an even number in \(C^7_1\) ways ,fill it with an even number in \(5^1\) ways and fill the remaining 6 digits with odd numbers in \(5^6\) ways. Again we need to subtract the numbers having 0 in the first digit location If first digit has a 0 , we can fill the remaining 6 digits with odd numbers in \(5^6\) ways.

Total no of 7 digit numbers whose sum of digits is even= \(5^7-5^6\)+\(C^7_5*5^7-C^6_4*5^6\)+ \(C^7_3*5^7-C^6_2*5^6\)+\(C^7_1*5^7-5^6\) =\(4*5^6+90*5^6+160*5^6+34*5^6\) =\(288*5^6\)

Zareentaj - What's the OA ?

Kudos for the patience displayed. Although if taken this approach in GMAT I wonder if it can be done in 2 mins? Assume Bunuel or Ian's approach is what they would be looking for... _________________

Re: How many different 7 digit members are their sum of whose [#permalink]

Show Tags

20 Aug 2014, 05:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: How many different 7 digit members are their sum of whose [#permalink]

Show Tags

02 Feb 2016, 10:55

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

Strategy, innovation, marketing, finance... The second module has been pretty engaging. Though, no lack of memorable times. There is no lack of high profile guest speakers. One of the...