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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.

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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 _________________

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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.

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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..

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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. _________________

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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?

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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. _________________

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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.

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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?

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
01 Jun 2011, 07:25

2

This post received KUDOS

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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)

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
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)

Re: How many four-digit odd numbers do not use any digit more than once? [#permalink]
27 Oct 2014, 07:47

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