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:

If the number begins with 8, there are \(5*8 = 40\) possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit).

If the number begins with 9, there are \(4*8 = 32\) possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit).

In all, there are \(40 + 32 = 72\) numbers that satisfy the constraints. The correct answer is C.

Now, when I calculate all the possible choices I come up with 81, meaning answer choice D.

Here is my reasoning:

All odd numbers in 800s - You have 9 choices for the second digit (0,1,2,3,4,5,6,7,9) - You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices = (5 x 4) + (4 x 5) = 40. (5 is for all the even and 4 is for all the odd)

All odd numbers in 900s - You have 9 choices for the second digit (0,1,2,3,4,5,6,7,8) - You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices = (5 x 5) + (4 x 4) = 41. (5 is for all the even and 4 is for all the odd)

for some reason my calculations don't match the explanations for the correct answer.

Here is the question:

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

(C) 2008 GMAT Club - m10#13

* 40 * 56 * 72 * 81 * 104

If the number begins with 8, there are \(5*8 = 40\) possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit).

If the number begins with 9, there are \(4*8 = 32\) possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit).

In all, there are \(40 + 32 = 72\) numbers that satisfy the constraints. The correct answer is C.

Now, when I calculate all the possible choices I come up with 81, meaning answer choice D.

Here is my reasoning:

All odd numbers in 800s - You have 9 choices for the second digit (0,1,2,3,4,5,6,7,9) - You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices = (5 x 4) + (4 x 5) = 40. (5 is for all the even and 4 is for all the odd)

All odd numbers in 900s - You have 9 choices for the second digit (0,1,2,3,4,5,6,7,8) - You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices = (5 x 5) + (4 x 4) = 41. (5 is for all the even and 4 is for all the odd)

Total: 40 + 41 = 81

Where I am going wrong??

Thanks.

You shouldn't count 9 in the units digit, it's already in hundreds

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

(C) 2008 GMAT Club - m10#13

* 40 * 56 * 72 * 81 * 104

C. Here is how I solved it, but other methods would be welcome.

For numbers between 800-899: Hundreds: one (i.e. 8) Tens: nine (i.e. 0-9, but not 8) Units: five (i.e. 1,3,5,7,9) 1x9x5=45 numbers; there are five numbers we need to exclude from this list: 811, 833, 855, 877, 899. So, between 800 and 899, there are 40 numbers that satisfy the constraints.

For numbers between 900-999: Hundreds: one (i.e. 9) Tens: nine (i.e. 0-8) Units: four (i.e. 1,3,5,7) 1x9x4=36 numbers; there are four numbers we need to exclude from this list: 911, 933, 955, 977. So, between 900 and 999, there are 32 numbers that satisfy the constraints.

Instead of getting confused with the odd number stuff just calculate the total number of three-digit integers greater than 800 such that all their digits are different.

Here is how we do it. _ _ _

the first place (hundreds place) can be filled by 8 or 9 ie 2 ways the second place can be filled in 9 ways (since one integer is already take we only have to choose from 9 instead of 10) the third place can be filled in 8 ways (since two integers are already take we have to choose from 8 digits)

hence total = 2 * 9 * 8 = 144.

Now there are equal number of odd and even numbers, so just 144/2 = 72 is the total number of odd numbers required.

Not a big fan of permutations! Anyway, answer is C

Multiply and add up your choices for each spot just like with a combination lock.

Hundreds Spot: 2 choices (8 & 9) Tens Spot: 9 choices (0,1,2,3,4,5,6,7 & [8 or 9]) Units Spot: 4 choices (1,3,5,7,9)... yes. there are 5 odd choices but you'd subtract one so the hundreds spot doesn't double up

Instead of getting confused with the odd number stuff just calculate the total number of three-digit integers greater than 800 such that all their digits are different.

Here is how we do it. _ _ _

the first place (hundreds place) can be filled by 8 or 9 ie 2 ways the second place can be filled in 9 ways (since one integer is already take we only have to choose from 9 instead of 10) the third place can be filled in 8 ways (since two integers are already take we have to choose from 8 digits)

hence total = 2 * 9 * 8 = 144.

Now there are equal number of odd and even numbers, so just 144/2 = 72 is the total number of odd numbers required.

Instead of getting confused with the odd number stuff just calculate the total number of three-digit integers greater than 800 such that all their digits are different.

Here is how we do it. _ _ _

the first place (hundreds place) can be filled by 8 or 9 ie 2 ways the second place can be filled in 9 ways (since one integer is already take we only have to choose from 9 instead of 10) the third place can be filled in 8 ways (since two integers are already take we have to choose from 8 digits)

hence total = 2 * 9 * 8 = 144.

Now there are equal number of odd and even numbers, so just 144/2 = 72 is the total number of odd numbers required.

possible first integers 2 (8 or 9) possible second integers 9 (one of 0 to 9 is taken as first integer) possible third integers 8 (two of 0 to 9 is already considered)

possible numbers = 2*8*9 = 144

as there are equal number of even and odd integers answer is 72 _________________

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

For 3-digit integers greater than 800, two series are there 800 and 900...

For number starting with 8...first digit is 8...second digit may have 9 choices (0-9,except 8) and third digit depends on second digit... For second digit as even (0,2,4,6)...numbers = 4*5 = 20 for second digit as odd(1,3,5,7,9)...numbers = 5*4 = 20

For odd three-digit integers greater than 800 and less than 900, the numbers satisfies the criteria are = 20+ 20= 40

For number starting with 9...first digit is 9...second digit may have 9 choices (0-8) and third digit depends on second digit... For second digit as even (0,2,4,6,8)...numbers = 5*4 = 20 (since there are four odd numbers only, i.e. 1,3,5,7) for second digit as odd(1,3,5,7)...numbers = 4*3 = 12 (since there are three odd numbers only)

For odd three-digit integers greater than 900 and less than 999, the numbers satisfies the criteria are = 20+ 12= 32

So, the total numbers satisfying the criteria are 40+32 = 72

I did the same way. But it took me more than 3 minutes to come up with the idea. Good question! _________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."