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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?

 Feb 8, 2023
 #1
avatar+2667 
+1

a)

 

Sam wrote 9 x 1 = 9 digits from 1 - 9

 

Sam wrote 90 x 2 = 180 digits from 10 - 99

 

So, he wrote a total of \(180 + 9 = \color{brown}\boxed{189}\) digits

 Feb 8, 2023
 #2
avatar+2667 
+1

b) The only digits that are 0 occur in 10, 20, 30, ... , 90

 

There are 9 of these, so the probability is \({9 \over 189} = \color{brown}\boxed{1 \over 21} \)

 Feb 8, 2023
 #3
avatar+2667 
+1

c) Note that each digit, except for 0, occurs 20 times - 10 times as a tens place and 10 times as a one place

 

So, the sum is just \(20(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = \color{brown}\boxed{900}\)

 Feb 8, 2023

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