Find all School-related info fast with the new School-Specific MBA Forum

It is currently 23 Nov 2014, 19:42

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How many four-digit odd numbers do not use any digit more than once?

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 20 Jul 2008
Posts: 4
Followers: 0

Kudos [?]: 1 [0], given: 0

How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 09 Jun 2009, 07:31
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

46% (02:28) correct 54% (01:15) wrong based on 52 sessions
How many four-digit odd numbers do not use any digit more than once?

A. 1728
B. 2160
C. 2240
D. 2268
E. 2520

[Reveal] Spoiler:
The answer is C, 2240.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.
[Reveal] Spoiler: OA
Director
Director
User avatar
Joined: 03 Jun 2009
Posts: 803
Location: New Delhi
WE 1: 5.5 yrs in IT
Followers: 65

Kudos [?]: 428 [0], given: 56

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 09 Jun 2009, 09:05
Answer should be D.

Let a 4-digit number be represented by ABCD

Here A can have any value between 1 to 9 - so total 9
B can have any value between 0 to 9, but not A - so total 9
C can have any value between 0 to 9, but not A or B - so total 8
D can have any value between 0 to 9, but not A, B or C - so total 7

No. of ALL possible 4-digit nos (without repeating any digit) = 9*9*8*7 = 4536
Half of these would be odd.
Therefor, no. of ODD possible 4-digit nos (without repeating any digit) = 4536 / 2 = 2268
_________________

ISB 2011-12 thread | Ask ISB Alumni @ ThinkISB
All information related to Indian candidates and B-schools | Indian B-schools accepting GMAT scores
Self evaluation for Why MBA?

1 KUDOS received
Intern
Intern
avatar
Joined: 20 Jul 2008
Posts: 4
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 09 Jun 2009, 16:00
1
This post received
KUDOS
Thank you all who responded.

The answer is C, 2240.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.
1 KUDOS received
Intern
Intern
avatar
Joined: 12 May 2009
Posts: 48
Location: Mumbai
Followers: 1

Kudos [?]: 3 [1] , given: 1

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 09 Jun 2009, 21:45
1
This post received
KUDOS
This question has to be solved reverse
ABCD is a 4 digit number

Since D (the unit digit) has to be odd, we have 5 choices => D=5
Starting from A;
A can hold 1 to 9, but cannot contain the digit in unit position, so choices will be 8 => A = 8
B can gold 0 to 9 (10 choices), but not the digits in A and D, so we ahve 8 options => B = 8
C can also hold 0 to 9 (10 choices), but not the digits in A, B and D, se we have 7 choices => C = 7

Calculating the total number of options = A*B*C*D = 8*8*7*5 = 2240

IMO C
Senior Manager
Senior Manager
User avatar
Joined: 15 Jan 2008
Posts: 295
Followers: 2

Kudos [?]: 22 [0], given: 3

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 10 Jun 2009, 06:17
I wonder as to why shud i not solve it the other around...


tha last digit can have 5 ways.
the second last digit can have 9 ways. ( 0-9) except the last digit.
the third last digit cab have 8 ways ( 0-9) except the last and the second last digit.
the fourth last digit can have 6 ways ( as no zero, last digiht, second last, thirdt last)

hence the total will be 45*48 = 2160 ways..

can anyone tell me from where am i missing the extra 80 ways..
Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3574
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 375

Kudos [?]: 1912 [0], given: 359

GMAT ToolKit User Premium Member
Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 10 Jun 2009, 08:33
Expert's post
Neochronic wrote:
I wonder as to why shud i not solve it the other around...


tha last digit can have 5 ways.
the second last digit can have 9 ways. ( 0-9) except the last digit.
the third last digit cab have 8 ways ( 0-9) except the last and the second last digit.
the fourth last digit can have 6 ways ( as no zero, last digiht, second last, thirdt last)

hence the total will be 45*48 = 2160 ways..

can anyone tell me from where am i missing the extra 80 ways..


The highlighted part is correct only if second, third and fourth digits don't equal zero. Otherwise, we will have 7 ways.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 78
Followers: 3

Kudos [?]: 14 [0], given: 2

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 05 Jan 2011, 07:21
EnergySP wrote:
Thank you all who responded.

The answer is C, 2240.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.


I know it is an old post but I received 2160 as an answer and can´t figure out which one is correct.

I start from the last digit:
5 possiblities for the last digit
9 possibilities or the 3rd digit
8 for the 2nd and
6 for the 1st

So we have 6*8*9*5=2160

But EnergySP solution is also correct. Now I am confused.
Can somebody help?
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 24025
Followers: 3730

Kudos [?]: 31114 [0], given: 3270

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 05 Jan 2011, 09:06
Expert's post
medanova wrote:
EnergySP wrote:
Thank you all who responded.

The answer is C, 2240.

Explanation from the book makes more sense now that I've thought about it a bit more.

Let ABCD be your four digits; D has to be odd since the four-digit number is odd (5 odd digits from the sequence). A has 8 options (1 thru 9, but one of the unit is reserved for digit D); B has 8 options aswell (0 thru 9, again one digit is reserved for digit D); C has 7 options (0 thru 9; one digit accounted for D and two A and B); D has 5 options since it must be odd.

8*8*7*5 = 2240

What threw me off was my assumtion that all four digits must be odd, but for a number to be odd only the last digit has to be odd. Hope this helps.


I know it is an old post but I received 2160 as an answer and can´t figure out which one is correct.

I start from the last digit:
5 possiblities for the last digit
9 possibilities or the 3rd digit
8 for the 2nd and
6 for the 1st

So we have 6*8*9*5=2160

But EnergySP solution is also correct. Now I am confused.
Can somebody help?


See Walker's post above: you'll have 6 choices for the 1st digit if 2nd or 3rd digit doesn't equal to zero, otherwise you'll have 7 choices. So this approach is not correct.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Take a Survey about GMAT Prep - Win Prizes!

Intern
Intern
avatar
Joined: 21 Sep 2010
Posts: 17
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 05 Jan 2011, 11:34
four Digit Number which is odd i.e it must end with 1,3,5,7 or 9


So if following 4 blanks represent the 4 digit number:

- - - -

Unit's place can be filled in by any of the 5 digits (1,3,5,7 or9)


so we get,


- - - 5

We are left with 9 other numbers to fill in the rest, but we cannot repeat and first digit cannot be zero otherwise the number will not be truly 4 digit number.

Hence,

8*8*7*5= 2240 (ans. C)
Intern
Intern
avatar
Joined: 12 May 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 01 Jun 2011, 07:07
Why can't we approach from the first number

1. Cannot be "0", hence number of options - 9
2. Leaving the number in 1st place, gives options - 9 again
3. Leaving the number in 1st/ 2nd place, gives options - 8
4. Last has to be an ODD number, so options -5

So numbers possible = 9X9X8X5 =3240
we do not have any options so the answer is definitely WRONG. Is it because the number of options that i am putting for last digit "5" is incorrect because some of them may have already been used up in 1/2/3 rd places.

So what should be the GENERIC order to giving the number of options possible in such problems?
2 KUDOS received
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 2037
Followers: 128

Kudos [?]: 984 [2] , given: 376

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 01 Jun 2011, 07:25
2
This post received
KUDOS
1
This post was
BOOKMARKED
CyberAsh wrote:
Why can't we approach from the first number

1. Cannot be "0", hence number of options - 9
2. Leaving the number in 1st place, gives options - 9 again
3. Leaving the number in 1st/ 2nd place, gives options - 8
4. Last has to be an ODD number, so options -5

So numbers possible = 9X9X8X5 =3240
we do not have any options so the answer is definitely WRONG. Is it because the number of options that i am putting for last digit "5" is incorrect because some of them may have already been used up in 1/2/3 rd places.

So what should be the GENERIC order to giving the number of options possible in such problems?


I think you should look for something called "Slot Method" in MGMAT guide for P&C.

The idea is to assign the number in ascending order of restriction.

Units place: 1,3,5,7,9- Total=5(Most restrictive)
Thousands place: No 0 and not the digit used by units place- Total=8 (Less restrictive)
Tens place: 2 digits used. Left:10-2=8 (Yet less restrictive)
Hundreds place: 3 digits used. Left:10-3=7(Yet less restrictive)

Total=5*8*8*7=2240

Ans: "C"
_________________

~fluke

Take a Survey about GMAT Prep - Win Prizes!

Intern
Intern
avatar
Joined: 12 May 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 01 Jun 2011, 15:42
fluke wrote:
CyberAsh wrote:
Why can't we approach from the first number

1. Cannot be "0", hence number of options - 9
2. Leaving the number in 1st place, gives options - 9 again
3. Leaving the number in 1st/ 2nd place, gives options - 8
4. Last has to be an ODD number, so options -5

So numbers possible = 9X9X8X5 =3240
we do not have any options so the answer is definitely WRONG. Is it because the number of options that i am putting for last digit "5" is incorrect because some of them may have already been used up in 1/2/3 rd places.

So what should be the GENERIC order to giving the number of options possible in such problems?


I think you should look for something called "Slot Method" in MGMAT guide for P&C.

The idea is to assign the number in ascending order of restriction.

Units place: 1,3,5,7,9- Total=5(Most restrictive)
Thousands place: No 0 and not the digit used by units place- Total=8 (Less restrictive)
Tens place: 2 digits used. Left:10-2=8 (Yet less restrictive)
Hundreds place: 3 digits used. Left:10-3=7(Yet less restrictive)

Total=5*8*8*7=2240

Ans: "C"
Thanks, much appreciated.
VP
VP
avatar
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1362
Followers: 12

Kudos [?]: 147 [0], given: 10

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 13 Jun 2011, 00:23
ABCD

D has 5 options
A has 8 options
B has 7 options + 1 option of 0.
C has 7 options

8*8*7*5
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

CEO
CEO
User avatar
Joined: 09 Sep 2013
Posts: 3233
Followers: 224

Kudos [?]: 44 [0], given: 0

Premium Member
Re: How many four-digit odd numbers do not use any digit more than once? [#permalink] New post 27 Oct 2014, 07:47
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 four-digit odd numbers do not use any digit more than once?   [#permalink] 27 Oct 2014, 07:47
    Similar topics Author Replies Last post
Similar
Topics:
10 Experts publish their posts in the topic How many four-digit positive integers can be formed by using gmihir 14 22 May 2012, 08:35
8 Experts publish their posts in the topic How many 4 digit even numbers do not use any digit more than rxs0005 17 28 Sep 2010, 09:03
How many four-digit numbers are there, if the three leftmost bmwhype2 6 01 Jan 2008, 22:43
How many odd numbers less than 5000 can be formed using the Ravshonbek 16 10 Nov 2007, 12:12
How many four-digit numbers that do not contain the digits 3 sperumba 5 17 Jan 2006, 17:09
Display posts from previous: Sort by

How many four-digit odd numbers do not use any digit more than once?

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.