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.

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:

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

I know this answer has been recently discussed but that explanation confuses me even more. ____________________________________________________________________________________ How many odd three-digit integers greater than 800 are there such that all their digits are different?

72= Answer

I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you.

I know this answer has been recently discussed but that explanation confuses me even more. ____________________________________________________________________________________ How many odd three-digit integers greater than 800 are there such that all their digits are different?

72= Answer

I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you.

First place = 8 or 9 = 2 Second place = 0 - 7 and either 8 or 9 = 8 Third place = 1, 3, 5, 7 and 9 but there could be odd integer in first or second place. so;

1. if 8 in first place and 4 even integers in second place = 1x4x5 = 20 2. if 8 in first place and one of 5 odds in second place and one of remaining 4 odds in third place = 1x5x4 = 20 3. if 9 in first place and one of 5 evens in second place and one of 4 odds in third place = 1x5x4 = 20 4. if 9 in first place and one of 4 odds in second place and one of remaining 3 odds in third place = 1x4x3 = 12

Hello, Can you please help me understand as how there are 8 digits in the middle - If first can be 8 then middle is 0,1,2,3,4,5,6,7,9 - which totals to 9. Can you please help

If first is 8 and last can be either of 3,5,7,9 (odd numbers only) so in middle it will have only 8 numbers left to be filled out of 0-9

From your choices above (then middle is 0,1,2,3,4,5,6,7,9 - which totals to 9), you have taken all the 4 odd numbers (3,5,7,9) at same time. Which cannot be possible. _________________

Constraints :- 1. Three digit number 2. Starting with 8 or 9 3. No repetition 4. Odd

When 8 is first digit:- 1. Second number even (0,2,4,6), third number (1,3,5,7,9)- total cases (4*5 =20) 2. Second number odd (1,3,5,7,9), third number (odd numbers -1=4) - total cases (5*4=20)

When 9 is first digit:- 3. Second number even (0,2,4,6,8), third number odd (1,3,5,7)- total cases (5*4 = 20) 4. Second number is odd (1,3,5,7), third number odd (odd numbers -1 =3) - total cases (4*3 = 12)

Total number of possible arrangements = 20+20+20+12 = 72

This is lengthy procedure and is risky because you may miss one case & answer will be completely different , IMO direct application of P&C theory will be better approach under time constraints . _________________

Sun Tzu-Victorious warriors win first and then go to war, while defeated warriors go to war first and then seek to win.

I hope my approach is correct anyway; Let N = total no of digits with different individual digits N= 2x9x8 = 144 no of even = no of odds = 72. (D) _________________

KUDOS me if you feel my contribution has helped you.

I just used slot method: 2 choices for the hundreds digit (8 or 9), 9 choices for the tens digit (0,1,2,3,4,5,6,7, 8 or 9), and 4 choices for the tens digit (1 or 9, 3 or 9, 5 or 9 , 7 or 9) = 2 x 9 x 4 = 72.

first digit can be selected from (8,9) in 2 ways second distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 9 ways since one digit is already selected Third distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 8 ways since two digits are already selected . So total number of ways digits can be slected is 2*8*9 But we need to consider odd number of digits so (2*8*9)/2=72 SO...ANSWER is C _________________

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!

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.

Total for this range: 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.

If hundreds digit is 8, there are 5 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*5=40 If hundreds digit is 9, there are 4 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*4=32 Total= 72

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!

good problem...every time i get such problems as these, i try to solve in conservative(old manual) method.....i should try to find the right method which is also the faster method....ANSWER is 72 and no doubt abt it. _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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

(A) 40 (B) 56 (C) 72 (D) 81 (E) 104

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!

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.

Total for this range: 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.

Total for this range: 1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hi Geek,

I wonder how u r solving all questions with proper explanations... HAts off for tat...

How u r solving by these strategy? Does any math strategy is there for this types of problem and could you pls help me to read those things....

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.

Total for this range: 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.

I am kicking myself for getting this problem wrong on account of a really bad silly mistake. I missed reading the "odd" in the question stem and ended up with getting an answer not in the given choices.

The good thing is that I knew how to approach it using the combinatrics method but on account of poor reading of the question stem I ended up choosing a wrong answer.

Bunuel's way of selecting the 3rd digit after selecting the first (8/9) was particularly helpful and quick.

Missed this question only because of not reading with full concentration. Real bad mistake. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

Hi everyone, most fastest way to solve this question is

8_ _, different odd no. will be 8 _ _. first digit is 8, for odd number third digit will be (1,3,5,7,9) i.e. 5 possibility 2nd digit we can fill from (0-9), since we are looking for different digit so choice for 2nd digit will 10(0-9) - 1st digit-2nd digit with 8 possibility 8 _(8) _(5) = 8*5 = 40

similarly for 9 _ _

for 3rd digit choices are {1,3,5,7} for 2nd digit choices are 10 - 1st digit-3rd digit = 8

so 8*4=32

so total = 72 different odd digits

Another simple wat to solve is b/w 800-900 we have 50 odd no. now we will substract the odd no. like {11,33,55,77,99}= 5 and 881,83,85,87,89 =5 i.e. 50-5-5 =40

b/w 900-999 we have 50 odd no substract {11,33,55,77,99}=5 and 991,93,95,97,99 = 5 and like 919,929 .....989=8 50-5-5-8 = 32 so different odd number will be 40+32 = 72