How many three-digit positive integers with different digits are odd?

With this method numbers like 121 and 755 are included. Hence where to start for the slot method is really important.

can you elaborate?

Hi bobnil


Take the following example.

When you say the first number has 9 options, it can be any number from 1 - 9. Say 1 comes in the first place.

Now the second slot can take any of the 8 remaining numbers, say 2

The last slot has to be an odd number where you have put 5 options. This includes 1. so we can get 121 as a possibility which is not what the question asks for.


In such cases start with the constraint, which is the last place, where you can have 5 options. Then the 1st place can only have 8 options (the number in the last place is not included, nor is 0). Then the second place, where the first and the last number is not included but 0 can come in, therefore 8 options.


Hope this helps

Arun Kumar

Awh! I missed the possibilities. Thank you very much.

TestPrepUnlimited and Bunuel,

_a__ * _b__ * __c_
If c=0,
there are 9 options in a and 8 in b: 8*9=72

If \(c\neq{0}\),
there are 4 options in c, 8 in a and 8 in b: 4*8*8=264

264+72=336

I think that we do need to divide in those 2 groups...

My mistake: the number is odd

Total three digit numbers: 999 -100 + 1 = 900

Total odd numbers out of 900 numbers: 450

Repetition allowed: 5 * 5 * 5 = 125

Remaining numbers: 450 - 125 = 325

Total '5' numbers have the same digits and are odd: 111, 333, 555, 777, 999

=> 325 - 5 = 320

Answer B

Lets calculate by parts

Odd numbers wherein ten's and hundred's digit are even - (EEO) = 4 x 4 x 5 = 80

Odd numbers wherein ten's and hundreds digit are odd- (OOO) = 5 x 4 x 3 = 60

Odd numbers wherein hundreds digit is even and tens digit is odd (EOO) = 4 x 5 x 4 = 80

Odd numbers wherein hundreds digit is odd and tens digit is even (OEO) = 5 x 5 x 4 = 100

Adding them all up, we get 320

We have a constraint that the three digit number is odd, thus only 1,3,5,7 and 9 can take units place.
Hundreds place can be taken by all except '0'. So, leaving aside '0' and units place digit we have 8 possibilities.
Tens place will not have hundreds place digit and units place digit, so it can have 0,1,2...9 but only 8 digits after placing digits at units and hundreds place.

Hence, total odd numbers are = 8*8*5 = 320.

Asked: How many three-digit positive integers with different digits are odd?

Let us have 3 digits in the form ___ ___ ___
Unit digit has 5 options = {1,3,5,7,9}
Hundredth digit has 8 options since 0 & unit digits are not available out of 10 digits.
Tenth digit will have 8 options since unit digit and hundredth digits are not available.

Total three-digit positive integers with different digits are odd = 5*8*8 = 320

IMO B

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.