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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)

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

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.

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

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

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
_________________

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.

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