How many strings of 8 digits end with an even digit? So there are $10$ combinations for each digit except the last which has 5 possibilities ($0,2,4,6,8$). Thus $10*10*10*10*10*10*10*5=50000000$ combinations right?
As a follow up, how many strings of 8 digits have at least one repeated digit?
I'm not sure how to approach this one. The first digit you have $10$ possibilities and then the next digit you only have $1$ possibility and for the next $6$ digits you have $10$ possibilities  for each.
 A: Yes, on the # of combinations with an even digit. 
For the case of strings with 8 digits with at least one repeated digit, try counting the total number of strings with 8 digits ($10^8$) and subtracting the number of strings with no repeated digits ($10*9*8*7*6*5*4*3$ where the number goes down by 1 for each place since you can't repeat any of the digits to the left of the digit you're placing, if you build the numbers left to right).
A: For the first question your reasing is almost correct, except that for the first place you have $9$ choices and not $10$ as an $8$ digit number can't begin with $0$.
For the follow up question, find the number of $8$ digit numbers with no repeated digits. (i.e. look at the complement).
A: There is a very simple way to answer your first question using basic logic:  
To start with, how many 8 digit numbers are there?
Eight digit numbers start at 10,000,000 and proceed to 99,999,999. That gives us a total of 90,000,000 possible numbers. Half of those numbers end with an even digit.
90,000,000 * 1/2 = 45,000,000.
So, that answer must be 45,000,000.
If instead, you want strings of eight digits, you can now include 00000000 through 99999999 which give us 100,000,000 possible strings. 
Still, half of those strings end with an even digit. That answer is 50,000,000.
