Question

If you calculate the following sum 9+99+999+9999+99999...where the last number to be added consists of nine digits of 9, how many times would the digit 1 appear in the answer?

Answer 1

Answer:

9 times

Step-by-step explanation:

9+99  = 108+999 = 1107+9999 = 11106.......... = 1111111101

Answer 2

Answer:

9 times

Step-by-step explanation:

Ones digit: 9*9=81 (Add 8)

Tens digit: 9*8+8=81 (Add 8)

Hundreds digit: 9*7+8=71 (Add 7)

Thousands digit: 9*6+7=61 (Add 6)

...

8th digit: 9*2+3=21

Last digit: 9*1+2=11

Therefore, by observing patterns, after each digit is added there is one '1'. There are 9 digits which is 9 '1's.