Question 418179
There are 10 choices for the 1st digit (0-9)
There are 10 choices for the 2nd digit (0-9)
There are 10 choices for the 3rd digit (0-9)
There are 10 choices for the 4th digit (0-9)
There are 10 choices for the 5th digit (0-9)
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So far, there are
{{{10*10*10*10*10 = 10^5}}} possible pin numbers
This must be multiplied by the number of
possible choices for the final digit which must be odd
It can't be 0 since the pin # would be even
It can be 1, 3, 5, 7, or 9
That is 5 choices, so there are 
5*10^5 possible 6-digit odd pin #'s