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?
+2 (a) To find the total number of digits Sam wrote ✏ , we need to add up the number of digits in each of the numbers from 1 to 99.
There are 9 one-digit numbers (1-9), which contribute 9 digits.
There are 90 two-digit numbers (10-99), each of which has 2 digits. So these numbers contribute 90 x 2 = 180 digits.
Therefore, the total number of digits Sam wrote is 9 + 180 = 189.
(b) The only digit that Sam wrote down that is a 0 is the digit in the number 10. Therefore, the probability that Sam chooses a 0 is 1/189.
(c) To find the sum of all the digits Sam wrote down, we can first find the sum of the digits in each of the numbers from 1 to 99.
For the one-digit numbers, the sum of the digits is simply the number itself. So the sum of the digits for the one-digit numbers is:
1 + 2 + 3 + ... + 9 = 45
For the two-digit numbers, we can split each number into its tens digit and its ones digit. The sum of the tens digits is:
1 + 2 + 3 + ... + 9 = 45
And the sum of the ones digits is:
1 + 2 + 3 + ... + 9 = 45
So the sum of the digits for the two-digit numbers is:
45 x 2 = 90
Therefore, the sum of all the digits that Sam wrote down is:
45 + 90 = 135
