Sam writes down the numbers 1, 2, 3, ..., 99
(a) How many digits did Sam write, in total?
(b) Sam chooses one of the digits written down, at random. What is the probability that Sam chooses a 0?
(c) What is the sum of all the digits that Sam wrote down?
A) The first 9 numbers only have 1 digit. So it's 9*1=9. Then the 90 other numbers have 2 digits, 90*2=180. Add those together, so there are 189 digits.
B) There are 9 numbers that end with 0, so 9 is the numerator. We know there are 189 total digits, so that is the denominator. So our fraction is 9/189, which is equal to 1/21.
C) First, the tens digits. Each tens digit repeats 10 times, so we do 10(1+2+3+4+5+6+7+8+9)=450. After that, we calculate the ones digits, which is 10 sets of 1+2+3+...+8+9=45, and 10*45=450. So, we add those two 450s together to get 900 as the sum of the digits.
