ruplun wrote:Hi,
how this math problem is solved with proper explanation...
1.How many odd three-digit integers greater than 800 are there such that all their digits are different?
Some test-takers might find it easier just to do a little counting and predict the rest.
The last digit has to be odd: 1, 3, 5, 7, or 9
The first digit has to be 8 or 9.
The middle digit has to be different from the first and the last.
Say the first digit is 8, the last 1.
For the middle digit we can use anything but 1 or 8, so we have 8 options: 0, 2, 3, 4, 5, 6, 7, 9
So if the first digit is 8 and the last is 1, we can form 8 acceptable numbers: 801, 821, 831, 841, 851, 861, 871, 891.
Since we have 5 options for the last digit, we have 5 * 8 = 40 good numbers in which the first digit is 8.
Using this logic, we'll get another 40 good numbers if the first digit is 9, except that we can't use 9 for the last digit (because then we'd have 2 9's), so we lose 8 of our good numbers. So if the first digit is 9, we'll get only 40-8=32 good numbers.
So there are 40+32 = 72 acceptable numbers.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
