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?
+6 A: This problem can be broken down into two parts.
Answer (a): 189
B: There are 9 numbers containing a zero (10, 20 ... 90). They each only have one zero therefore there are 9 zeros. The total number of digits is 189 so you have 9/189 which can be simplified to 1/21
Answer (b): 1/21
C: Each digit appears once in the single-digit numbers. Then for the next 8 "tens" they appear once in 7 of the "tens" and 9 times in one "tens" category. For example, 9 appears once in the 10's (19) but appears 9 times in the 90s (91, 92...). Therefore, each digit appears 17 times. So you have (17*1)+(17*2)...(17*9) which should amount to 765.
Answer (c): 765
