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 odd three-digit integers greater than 800 are there such that all their digits are different?

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.
_________________

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

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

09 Oct 2008, 10:54

IanStewart wrote:

vr4indian wrote:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.

Ian why are we computing them separately ? can we not do them together ?

Thanks
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

09 Oct 2008, 11:53

vr4indian wrote:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40 56 72 81 104

Assume: X = {x | 800 < x < 1,000 and all x's digits are different}

Set of number = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Possible digits for x = 1, 3, 5, 7, and 9

We have to separate into two cases.

Case 1: 800 < x < 899 ---------- because the first digit is 8 or an even number, there are 5 odd number available for the last digit. Possible numbers = 1 x 8 x 5 = 40 numbers

Case 2: 900 < x < 1,000 ---------- because the first digit is 9 or an odd number, there are only 4 odd numbers available for the last digit. Possible numbers = 1 x 8 x 4 = 32 numbers

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

23 Jan 2010, 01:19

IanStewart wrote:

vr4indian wrote:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

27 Nov 2011, 21:36

1

This post received KUDOS

Can we do it like this ? _ _ _ 2 Possibilities for 100th digit 8 and 9, 9 Possibilities for 10th Digit, 8 Possibilities for Unit digit. that gives us 2 * 9 * 8 = 144, Since we need only odd numbers 144/2 = 72. And thats the right answer. Can anyone tel me if i am missing something or this method is alright ?

Can we do it like this ? _ _ _ 2 Possibilities for 100th digit 8 and 9, 9 Possibilities for 10th Digit, 8 Possibilities for Unit digit. that gives us 2 * 9 * 8 = 144, Since we need only odd numbers 144/2 = 72. And thats the right answer. Can anyone tel me if i am missing something or this method is alright ?

Yes, you are missing something and there is a reason why Ian did what he did. Take numbers in the range 500 - 600. How many numbers are there such that all 3 digits are different? 1 (hundred's digit) * 9 (ten's digit) * 8 (unit's digit) = 72 How many of them are odd? 5 is odd so number of possibilities for unit's digit is 4. [highlight]Number of odd numbers = 1 * 8 * 4 = 32[/highlight] Number of even numbers = 72 - 32 = 40

Now take numbers in the range 800 - 900. Total numbers where all digits are different = 72 (as before) [highlight]Number of odd numbers = 1 * 8 * 5 = 40[/highlight] (now there are 5 possibilities for the unit's digit) Number of even numbers = 72 - 40 = 32

So the number of even and odd numbers are not half of the total numbers in each range. You see why above. The number of odd numbers depends on the range (whether the hundred's digit is odd or even).

The reason you still got your answer is that the range is 800 - 1000 which has the 800 - 900 and the 900 - 1000 range in it. It averaged out to be half because in the first range 40 numbers are odd and in the second range 32 are odd so they averaged out to 36 i.e. half of 72.

If instead the question says how many 3 digit numbers above 700 are odd with all digits different, the answer will be 104, not 108. Your method will not work in this case. So be careful.
_________________

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

28 Nov 2011, 05:04

1

This post received KUDOS

Thanks Karishma, Actually i missed out the "ODD Numbers" part in the question earlier and my answer was 144, but none of the answers were matching. Then I saw 72 which is equal to 144/2 so i thought its okay to do this way. Now I understand it better

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

28 Nov 2011, 08:36

1

This post received KUDOS

separate odd and even integers (0,1,2,4,6,8) and (1,3,5,7,9) if the 100's place is 8 then the number of ways 1*4(all even except 8)*5(all odd) + 1*5(all odd)*4(four odds left to be chosen)=20+20 =40 if the 100's place is 9 1*5(all even)*4(all odd except 9) + 1*4(all odd)*3(three odd left)=20+12=32

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

09 Dec 2011, 11:14

VeritasPrepKarishma wrote:

arjunbt wrote:

Can we do it like this ? _ _ _ 2 Possibilities for 100th digit 8 and 9, 9 Possibilities for 10th Digit, 8 Possibilities for Unit digit. that gives us 2 * 9 * 8 = 144, Since we need only odd numbers 144/2 = 72. And thats the right answer. Can anyone tel me if i am missing something or this method is alright ?

Yes, you are missing something and there is a reason why Ian did what he did. Take numbers in the range 500 - 600. How many numbers are there such that all 3 digits are different? 1 (hundred's digit) * 9 (ten's digit) * 8 (unit's digit) = 72 How many of them are odd? 5 is odd so number of possibilities for unit's digit is 4. [highlight]Number of odd numbers = 1 * 8 * 4 = 32[/highlight] Number of even numbers = 72 - 32 = 40

Now take numbers in the range 800 - 900. Total numbers where all digits are different = 72 (as before) [highlight]Number of odd numbers = 1 * 8 * 5 = 40[/highlight] (now there are 5 possibilities for the unit's digit) Number of even numbers = 72 - 40 = 32

So the number of even and odd numbers are not half of the total numbers in each range. You see why above. The number of odd numbers depends on the range (whether the hundred's digit is odd or even).

The reason you still got your answer is that the range is 800 - 1000 which has the 800 - 900 and the 900 - 1000 range in it. It averaged out to be half because in the first range 40 numbers are odd and in the second range 32 are odd so they averaged out to 36 i.e. half of 72.

If instead the question says how many 3 digit numbers above 700 are odd with all digits different, the answer will be 104, not 108. Your method will not work in this case. So be careful.

I am definitely taking more than the prescribed 2 minutes to solve these questions. Am i missing something ?

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

09 Dec 2011, 11:33

1

This post received KUDOS

WE have following slots: ABC(for 800) and XYZ (for 900) A is reserved for 8 and C has to be odd. X is reserved for 9 and Z has to be odd, But Z cannot be 9. Therefore, For A and X we have 1 choice. For C we have 5 choices. For Z we have 4 Choices. For B and Y we have 8 Choices. Therfore we have: (1C1*8C1*5C1)+(1C1*8C1*4C1) 40+32=72

I am definitely taking more than the prescribed 2 minutes to solve these questions. Am i missing something ?

There is nothing called the 'prescribed two minutes'. For the lower level question, you will take less than two mins and for the higher level ones, you will take more than that. Two minutes is just the average. Also, once you work on quite a few such questions, you will be able to do them in 2 minutes or less.
_________________

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

12 Dec 2011, 23:41

1

This post was BOOKMARKED

OA is C 72 800 to 900: choice for the unit digit place must be odd(1,3,5,7,9) is 5 choices for the tens digit (exclude 8 and odd digit used at unit place) 8 choices

5*8 = 40 choices

From 900 to 1000: choice for the unit digit place must be odd(1,3,5,7,) is 4 choices for the tens digit (exclude 9 and odd digit used at unit place)8

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

05 Apr 2012, 11:49

IanStewart wrote:

vr4indian wrote:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.

Understood the concept here, but i just cannot understand one thing, regarding tens digit why do we have 8 choices? as far my understanding goes for 800 its 8(hundred digit) 0,2,4,6( 4 digits available for tens) as we cannot have any odd digit here else it will coincide with unit digit. and 1,3,5,7,9(five available for unit digit) can someone make me understand here?
_________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back!

How many odd three-digit integers greater than 800 are there such that all their digits are different?

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.

Understood the concept here, but i just cannot understand one thing, regarding tens digit why do we have 8 choices? as far my understanding goes for 800 its 8(hundred digit) 0,2,4,6( 4 digits available for tens) as we cannot have any odd digit here else it will coincide with unit digit. and 1,3,5,7,9(five available for unit digit) can someone make me understand here?

We can have odd digit for tens digit, though it must be different from that we used for units digit.

Next, since all 3 digit must be distinct then if you use one for hundreds and one for units then there are 8 choices left for tens digit.

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...