vinay1983 wrote:How many positive integers between 1 and 1000 contain the number "3" as at least one of their digits?

To make it easier on ourselves, let's consider the positive integers from 0 to 999 inclusive. As you can see, the answer will still be the same. All we have done is replaced the 1000 with 0, and neither of these have a "3"

IMPORTANT: When it comes to questions involving "at least," it's often a good idea to use the version of the complement rule that we use in probability.

That is: (# of integers with zero 3) + (# of integers with at least one 3) = total # of integers
There are 1000 integers from 0 to 999 inclusive.
So, we can write: (# of integers with zero 3's) + (# of integers with at least one 3) = 1000


# of integers with zero 3's
First take the digits from 0 to 9 and remove the "3"
This leaves us with 9 integers.
Now let's "build" some integers from 000 to 999.
First, there are 9 options for the hundreds digit.
There are 9 options for the tens digit.
There are 9 options for the units digit.
So, in total, the number of integers with zero 3's = 9 x 9 x 9 = 729


We get: (729) + (# of integers with at least one 3) = 1000
So, # of integers with at least one 3 = 271

Answer: 271

Cheers,
Brent

mainbhidhruv wrote:@Brent In your Solution above you are considering 000 as a number when in question the range is 1 to 1000, why is it so- Please clarify .

Since, there are no three's in 1000, we could just ask "Among the integers from 1 to 999, how many have at least one 3?"
Likewise, there there are no three's in 000, we could just ask "Among the integers from 000 to 999 inclusive, how many have at least one 3?"
This is what I did.
I recognized that there are 1000 integers from 000 to 999 inclusive
Then I calculated there are 729 integers with zero 3's
So, the other 271 must have at least one 3.

mainbhidhruv wrote: When I calculate 3 digit numbers with no 3's
1 digit : 8 (1,2,4,5,6,7,8,9 )
2 digit : 8*9 = 72
3 digit : 8*9*9 = 648

8+72+648 = 728

Where am I missing 1 number, Please correct me.

You're almost there.
Since you don't seem to be including 1000 in your calculations, you are essentially answering the question, "Among the integers from 1to 999 inclusive, how many have at least one 3?"
Since there are 999 integers from 1 to 999 inclusive, we must subtract 728 from 999 to get 271

Same answer, different approach.

Cheers,
Brent