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# counting

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6

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?

Mar 25, 2023

#1
+1

(a) 99 digits in total

(b) 0 because Sam didn't write the number 0

(c) 4950

Hope this helps ^^

Mar 25, 2023
#2
+1

a) Sam has total digits of 189

b) The possibility is 9/189; as the numbers 10, 20, 30 ...90 have the number 0.

c) The sum of n = n(n+1)/ 2

-> 99(99 + 1) / 2

-> 99(100) / 2

-> 9900 / 2

-> 4950

Mar 25, 2023
#3
0

For (c) I think you are supposed to add up ALL the digits he wrote down like this:

1+2+3+4+5+6+7+8+9+1+0(for ten)+1+1(for eleven)+1+2(for twelve).......and so on. You should get a total of 900 for the the sum of 189 digits he wrote down.

Guest Mar 25, 2023
#5
0

A)99 digits for sam's

B)0 because Sam did not include a zero

C)4934 is the total of sam's written digits

Mar 25, 2023