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:
How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40 B. 60 C. 72 D. 81 E. 104
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 the 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 Bunuel,
I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.
How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40 B. 60 C. 72 D. 81 E. 104
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 the 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 Bunuel,
I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.
801, 803, 805, 807, ... are all odd integers.
_________________
Thanks for your explanation! Is there an alternative method to solve this question? trying to think how could this problem be solve by applying the combinatorics formula with repeating elements [N!][/A!B!]
Thanks for your explanation! Is there an alternative method to solve this question? trying to think how could this problem be solve by applying the combinatorics formula with repeating elements [N!][/A!B!]
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't get explanation for the second digit part can you please list out all possibilities ?
3 digit integers such that all three digits are different between 800-899 : 100th place is reserved for 8, 10th place can be taken by any of the remaining 9 digits (0,1,2,3,4,5,6,7,9) Unit's pace can be taken by any of the remaining 8 digits so 9*8 = 72 such numbers Similary there are 72 such numbers between 900-999 Total number 72 + 72 =144 Between 800-999 (both inclusive) there are 200 numbers, so there should be equal number of odd and even numbers: hence total odd numbers = 144/2 = 72
I dont get the explanation for the 2nd digit. 800-899 1st digit: 1 choice (8) 3rd digit: 5 choices (1,3,5,7,9) 2nd digit: 10 digits - 1 digit - 5 digits (0,2,4,6)
This way, you cannot get 811. If I use your solution above, it does not exclude 811, 822, 833, 844 etc.
The first step when solving a question ALWAYS must be reading CAREFULLY: How many odd three-digit integers greater than 800 are there such that all their digits are different?
_________________
The first step when solving a question ALWAYS must be reading CAREFULLY: How many odd three-digit integers greater than 800 are there such that all their digits are different?
Hi Bunuel,
I understand the question stem to explicitly exclude cases where some of the 3 digits are the same, i.e. numbers such as 811, 822, 833 should be excluded. That is why I have less choices for the 2nd digit in my solution as you have. That way I ensure that none of the 3 digits are the same. Could you please elaborate on this?
The first step when solving a question ALWAYS must be reading CAREFULLY: How many odd three-digit integers greater than 800 are there such that all their digits are different?
Hi Bunuel,
I understand the question stem to explicitly exclude cases where some of the 3 digits are the same, i.e. numbers such as 811, 822, 833 should be excluded. That is why I have less choices for the 2nd digit in my solution as you have. That way I ensure that none of the 3 digits are the same. Could you please elaborate on this?
So, you say that it should be - In the range 800 - 900: 1*5*4? That's not right. For the 2nd digit you can use any digit apart from those used for the first and the thrid one, so the number of options is 10 - 1 - 1 = 8.
If still not convinced check other ways of solving HERE.
_________________
Think of the digits as the menu choices for a three-course meal: Appetizers (hundreds digit), Main (tens digit), and Desserts (units digit)
Appetizer menu has 2 choices: {8,9} -- eat "9" Main menu has 3 choices: {0,1,2,3,4,5,6,7,8} i.e. "9" is already eaten Dessert menu has 4 choices: {1,3,5,7} i.e. again, "9" is already eaten
Total three-course meal possibilities = 2*9*4 = 72
It took me a little time to understand the question. Like, I had to read it thrice to formulate a concept in my head but once that was done, it was an easy ride.
_________________
close
52% (01:33) correct 
