Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 07 Aug 2010
Posts: 1

How many different 7 digit members are their sum of whose
[#permalink]
Show Tags
07 Aug 2010, 06:10
Question Stats:
33% (00:19) correct 67% (00:39) wrong based on 7 sessions
HideShow timer Statistics
How many different 7 digit members are their sum of whose digits is even ?




Math Expert
Joined: 02 Sep 2009
Posts: 49381

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 14:57
I guess the question is: How many different 7digit 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 7digit integers: 1,000,0001,000,009  half has even sum and half odd; Next ten 7digit integers: 1,000,0101,000,019  also half has even sum and half odd; And so on. There are \(9*10^6\) 7digit integers, half of it, \(45*10^5\) will have odd sum of their digits, or as crack700 wrote \(288*5^6\), (\(288*5^6=2^5*9*5^6=(9*5)*(2^5*5^5)=45*10^5\)). Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 367
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 10:13
My attempt: 7 digit number is of the form abcdefg. In this the sum of the digits would end up as a even number only if the following is true. a) There number of even number among these numbers  a, b, c, d, e, f and g should be either 1 or 3 or 5 or 7. Also the number of even and odd single digit numbers are 5. (1,3,5,7,9 and 0, 2, 4, 6, 8) To summarize, let us take the even digit to be e and odd digit to be o. Hence the number abcdefg could be either ooooooe OR ooooeee OR ooeeeee OR eeeeeee Number of ways to choose a even or odd digit is 5. Hence answer is 4*5^7.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog



Manager
Joined: 20 Mar 2010
Posts: 80

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 13:58
zareentaj wrote: 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 = \(5^75^6\) 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 = \(C^7_5*5^5*5^2C^6_4*5^4*5^2\) 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 = \(C^7_3*5^3*5^4C^6_2*5^2*5^4\) 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 = \(C^7_1*5^1*5^65^6\) 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= \(5^75^6\)+\(C^7_5*5^7C^6_4*5^6\)+ \(C^7_3*5^7C^6_2*5^6\)+\(C^7_1*5^75^6\) =\(4*5^6+90*5^6+160*5^6+34*5^6\) =\(288*5^6\) Zareentaj  What's the OA ?
_________________
___________________________________ Please give me kudos if you like my post



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1344

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 15:46
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



Director
Status: Apply  Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 636
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 16:37
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^(N1)/2?
_________________
Consider kudos, they are good for health



Director
Status: Apply  Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 636
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 16:41
crack700 wrote: zareentaj wrote: 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 = \(5^75^6\) 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 = \(C^7_5*5^5*5^2C^6_4*5^4*5^2\) 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 = \(C^7_3*5^3*5^4C^6_2*5^2*5^4\) 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 = \(C^7_1*5^1*5^65^6\) 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= \(5^75^6\)+\(C^7_5*5^7C^6_4*5^6\)+ \(C^7_3*5^7C^6_2*5^6\)+\(C^7_1*5^75^6\) =\(4*5^6+90*5^6+160*5^6+34*5^6\) =\(288*5^6\) 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...
_________________
Consider kudos, they are good for health



Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 367
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: How to solve this question
[#permalink]
Show Tags
07 Aug 2010, 17:09
Wow! . Thanks Bunuel/IanStewart/crack700 for the explanation.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog



Math Expert
Joined: 02 Sep 2009
Posts: 49381

Re: How to solve this question
[#permalink]
Show Tags
08 Aug 2010, 01:39



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India

Re: How many different 7 digit members are their sum of whose
[#permalink]
Show Tags
10 May 2017, 09:45
zareentaj wrote: How many different 7 digit members are their sum of whose digits is even ? You can think of it in terms of subsequent number changes. The first 7 digit number is 1,000,000. The sum of the digits will be 1, an odd number. The next 7 digit number is 1,000,001. Its units digit will be 1 more than the previous so the sum of its digits will increase by 1 and hence be even. and so on the sum will alternate between odd and even till after the next 8 numbers when the tens digit changes and we get 1,000,010. This sum is even and now again each subsequent number will increase by 1 and the sum of digits will alternate again between odd and even for the next 9 numbers. So 5 of them will be even. And so on... So exactly half the numbers will have even sum and half will have odd sum. Total 7 digit numbers = 9 * 10*10*10*10*10*10 = 9,000,000 (The first digit cannot be 0. Rest all digits can take any value) Half of them will have odd sum of digits. So 4,500,000
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



NonHuman User
Joined: 09 Sep 2013
Posts: 8160

Re: How many different 7 digit members are their sum of whose
[#permalink]
Show Tags
26 Jun 2018, 07:19
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: How many different 7 digit members are their sum of whose &nbs
[#permalink]
26 Jun 2018, 07:19






