M10-13

Notice that there are 5 choices for units digit when the hundreds digit is 8 and 4 choices when the hundreds digit is 9. So you should count numbers in these 2 ranges separately.

tmukogo,

I tried to do in this way:
think first considering all integers with all different algorithms, odd and even.
For the first digit we have: 8 or 9 = two possibilities.
For the second, we can have all numbers between 0 and 9, except the "8" or "9" digit chosen so: 10-1=nine.
For the third: the same reasoning except the two previous digit already taken: 10-2=eight.
two*nine*eight=144. We divide it per two to consider only the odd numbers: 144/2=72.

obs:We could have had a problem with "start" and "end" numbers using this reasoning then the answer could be 72-1, for example. But I find it difficult to consider this xD

I am confused by the second digit logic.

I went figured 10 figures available to the second digit. And then got rid of any that were included in the two other digits (i.e. for the 800's: I got rid of 8, then all odd digits.. thus 4 remain). This answer got me to 40.

I want to clarify, the reason that my logic was flawed was that I excluded all odd figures in the second digit; thus I excluded far too many numbers. For instance, I would have excluded 871 simply because 7 is an odd number that is capable of appearing in the third digit.

Following this logic, the reason that the second digit has 8 possibilities is because you start with 10 and subtract the 8 (first digit) and odd digit that will appear in the third digit.


Thanks! Want to make sure I am thinking of this correctly.

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Yes, that;s correct.

I started with a unit digit.

Unit digit can have [1,3,5,7,9]

=> if a unit digit is 9, then the number of ways = 1 (unit digit) x 1 (hundredth digit) x 8 (tenth digit) = 8
=> if a unit digit is not 9, then the number of ways = 4 (unit digit) x 2 (hundredth digit) x 8 (tenth digit) = 64

Answer = 8 +64 = 72

Duplicate of M25-20. Unpublished....................

In these "counting" kind of questions, I tend to answer correctly but waste more than 2 mins (in the range of 3/4). Does anyone have any tips to speed up "counting" type of problems?