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:
If the number begins with 8, there are possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit). If the number begins with 9, there are possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit). In all, there are numbers that satisfy the constraints.
The correct answer is C.
GMAC25 : i think there are 9 possiblities for middle digit, 0-9 and excluding 8, now how can we say that there are 5 possibilties for last digit when middle digit is 1 then possible last digit values are 3,5,7,9.
Last digit is always ODD: case1: first digit is 8, then last digit has 5 options(1,3,5,7,9) and middle digit has 8 options as we have already chosen two digits for the first and the last positions. total numbers = 1*8*5 = 40
case2: first digit is 9, then last digit has 4 options(1,3,5,7) and middle digit has 8 options again. total numbers = 1*8*4 = 32 Ans:72
NOTE: last digit should always be one of 1,3,5,7,9
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.
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.
Re: How many odd three-digit integers greater than 800 are there
[#permalink]
Show Tags
22 May 2017, 21:54
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
61% (01:42) correct 
