Last visit was: 19 Nov 2025, 02:34 It is currently 19 Nov 2025, 02:34
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
EnergySP
User avatar
Joined: 20 Jul 2008
Last visit: 15 Mar 2010
Posts: 4
Own Kudos:
23
 [20]
Posts: 4
Kudos: 23
 [20]
How many four-digit odd numbers do not use any digit more than once? 09 Jun 2009, 08:31
2
Kudos
Add Kudos
18
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

57% (01:55) correct 43% (02:06) wrong based on 394 sessions

History

Date
Time
Result
Not Attempted Yet
How many four-digit odd numbers do not use any digit more than once?

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

Show 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.
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
fluke
User avatar
Retired Moderator
Joined: 20 Dec 2010
Last visit: 24 Oct 2013
Posts: 1,099
Own Kudos:
5,095
 [10]
Given Kudos: 376
Posts: 1,099
Kudos: 5,095
 [10]
How many four-digit odd numbers do not use any digit more than once? 01 Jun 2011, 08:25
7
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
CyberAsh
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?
Show more

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"
General Discussion
User avatar
bigoyal
User avatar
Joined: 03 Jun 2009
Last visit: 08 Jul 2011
Posts: 577
Own Kudos:
2,394
Given Kudos: 56
Location: New Delhi
Concentration: IT Consultancy
WE 1: 5.5 yrs in IT
Posts: 577
Kudos: 2,394
How many four-digit odd numbers do not use any digit more than once? 09 Jun 2009, 10:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
EnergySP
User avatar
Joined: 20 Jul 2008
Last visit: 15 Mar 2010
Posts: 4
Own Kudos:
23
 [3]
Posts: 4
Kudos: 23
 [3]
How many four-digit odd numbers do not use any digit more than once? 09 Jun 2009, 17:00
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
vesam
User avatar
Joined: 12 May 2009
Last visit: 11 Dec 2009
Posts: 16
Own Kudos:
7
 [3]
Given Kudos: 1
Location: Mumbai
Posts: 16
Kudos: 7
 [3]
How many four-digit odd numbers do not use any digit more than once? 09 Jun 2009, 22:45
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
Neochronic
User avatar
Joined: 15 Jan 2008
Last visit: 05 Jul 2010
Posts: 131
Own Kudos:
68
Given Kudos: 3
Posts: 131
Kudos: 68
How many four-digit odd numbers do not use any digit more than once? 10 Jun 2009, 07:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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..
User avatar
walker
User avatar
Joined: 17 Nov 2007
Last visit: 25 May 2025
Posts: 2,398
Own Kudos:
10,717
 [1]
Given Kudos: 362
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Expert
Expert reply
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,398
Kudos: 10,717
 [1]
How many four-digit odd numbers do not use any digit more than once? 10 Jun 2009, 09:33
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Neochronic
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..
Show more

The highlighted part is correct only if second, third and fourth digits don't equal zero. Otherwise, we will have 7 ways.
User avatar
medanova
User avatar
Joined: 19 Aug 2010
Last visit: 15 Feb 2012
Posts: 51
Own Kudos:
105
Given Kudos: 2
Posts: 51
Kudos: 105
How many four-digit odd numbers do not use any digit more than once? 05 Jan 2011, 08:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EnergySP
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.
Show more

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?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,379
Own Kudos:
778,172
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,379
Kudos: 778,172
How many four-digit odd numbers do not use any digit more than once? 05 Jan 2011, 10:06
Kudos
Add Kudos
Bookmarks
Bookmark this Post
medanova
EnergySP
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?
Show more

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.
avatar
denver
avatar
Joined: 21 Sep 2010
Last visit: 17 Apr 2011
Posts: 10
Own Kudos:
5
Posts: 10
Kudos: 5
How many four-digit odd numbers do not use any digit more than once? 05 Jan 2011, 12:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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)
avatar
CyberAsh
avatar
Joined: 12 May 2011
Last visit: 13 Sep 2019
Posts: 2
Posts: 2
Kudos: 0
How many four-digit odd numbers do not use any digit more than once? 01 Jun 2011, 08:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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?
avatar
CyberAsh
avatar
Joined: 12 May 2011
Last visit: 13 Sep 2019
Posts: 2
Posts: 2
Kudos: 0
How many four-digit odd numbers do not use any digit more than once? 01 Jun 2011, 16:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
fluke
CyberAsh
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"
Show more
Thanks, much appreciated.
avatar
Glengubbay1
avatar
Joined: 06 Oct 2014
Last visit: 15 Oct 2016
Posts: 5
Own Kudos:
11
Given Kudos: 24
Concentration: Finance
Schools: HBS '19 Wharton '19 CBS '19
GMAT 1: 700 Q45 V41
GPA: 3.8
Schools: HBS '19 Wharton '19 CBS '19
GMAT 1: 700 Q45 V41
Posts: 5
Kudos: 11
How many four-digit odd numbers do not use any digit more than once? 05 Nov 2015, 12:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Just curious how do you know where the option for zero can be counted as well? i.e hundreds vs thousands? I would assume that as long as it is not counted in the unit digits it's all that matters.

It's quite clear to me why not the digits because then for example 7530 becomes divisible by 2. But what difference does 7053 and 7503 make?

Thank you!
User avatar
shashankism
User avatar
Joined: 13 Mar 2017
Last visit: 23 Dec 2024
Posts: 609
Own Kudos:
694
Given Kudos: 88
Affiliations: IIT Dhanbad
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE:Engineering (Energy)
Posts: 609
Kudos: 694
How many four-digit odd numbers do not use any digit more than once? 23 Jul 2017, 04:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EnergySP
How many four-digit odd numbers do not use any digit more than once?

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

Show 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.
Show more

The question asks for 4 digit odd number which do not use any digit more than once.

So, the questions is basically asking to arrange the digits from 0,1,2,..,3,9 without repetition to form a 4 digit number with last digit as 1,3,5,7 or 9 only.

Lets say the 4 digit number be PQRS.
So, S can take numbers as 1,3,5,7,9 i.e. 5 ways.
Now we will chk nos. for P as it has restriction of being a non - zero. So it can take numbers as 1,2,..,9 - the no. taken in S which can be done in 8 ways,
Now, Q can take nos. 0,1,2,..., 9 minus the no. taken in S & P which can be done in 8 ways.
Now, R can take nos 0,1,2,...9 minus the no. taken in S, P & Q which can be done in 7 ways.

So, total count of such numbers = 8*8*7*5 = 2240

Answer C
User avatar
sarathgopinath
User avatar
Joined: 22 Aug 2016
Last visit: 04 Feb 2025
Posts: 68
Own Kudos:
102
Given Kudos: 77
Location: India
Concentration: Operations, Finance
GMAT 1: 650 Q49 V29
GMAT 2: 710 Q49 V37
GPA: 3.5
WE:Other (Education)
Products:
My Reviews
GMAT 2: 710 Q49 V37
Posts: 68
Kudos: 102
How many four-digit odd numbers do not use any digit more than once? 31 Jul 2017, 06:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
bigoyal
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
Show more

I got the right answer but I'm not able to understand why this method is wrong.
please help Bunuel
User avatar
firas92
User avatar
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Own Kudos:
1,725
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
My Reviews
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
Posts: 616
Kudos: 1,725
How many four-digit odd numbers do not use any digit more than once? 03 Aug 2019, 02:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
We have 5 options for the units digit (1,3,5,7 and 9)

We have 8 options for the thousands digit (1-9 except the digit used in the units place)

We have 8 options for the hundreds digit (0-9 except the digits used in units and thousands places)

We have 7 options for the tens digit (0-9 except the digits used in units, thousands and hundreds places)

So total number of four-digit odd numbers do not use any digit more than once = 8*8*7*5 = 2240

Answer is (C)

Posted from my mobile device
User avatar
exc4libur
User avatar
Joined: 24 Nov 2016
Last visit: 22 Mar 2022
Posts: 1,684
Own Kudos:
1,447
Given Kudos: 607
Location: United States
Posts: 1,684
Kudos: 1,447
How many four-digit odd numbers do not use any digit more than once? 03 Aug 2019, 08:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
DarkHorse2019
How many four-digit odd numbers do not use any digit more than once?

(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520

Source: Math Bible (Jeff Sacksmann)
Show more

given: there are total of 10 digits (0,1,2,3,4,5,6,7,8,9),
and odd numbers cannot have an even units digit, so there are 5 digits/choices (1,3,5,7,9).

units: 5 choices (because odd numbers cannot have even units digits)
thousands: 8 choices (10-1-1=8 choices because you can't have 0 and we used a digit for the units)
hundreds: 8 choices (10-2=8 digits left, since we used two digits for the thousands and units)
tens: 7 choices (10-3=7 digits left, since we used three digits for the thousands, hundreds and units)

total four-digit odd integers with different digits: 5*8*8*7=35*64=2240.

Answer (C).
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,583
Own Kudos:
1,079
Posts: 38,583
Kudos: 1,079
How many four-digit odd numbers do not use any digit more than once? 16 Jul 2024, 12:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Bunuel
Math Expert
105379 posts
Krunaal
Tuck School Moderator
805 posts