Let A = 1/1.00...001 (there are 99 zeros in the expression). Convert A in decimal number and describe the pattern after the decimal point.

-> If there was zero 0s in a row, then A would be equal to:

1/1.1= 0.909090.....

-> If there was one 0 in a row, then A would be equal to:

1/1.01 = 0.9900990099......

-> If there was two 0s in a row, then A would be equal to:

1/1.001 = 0.999000999000999...........

.....................................

......................................

If there were n zeros in a row, then A would be:

1

..................... = 0.99........9900..........0099........99

1.00......001 n+1 9s n+1 0s n+1 9s

n zeros

Substituting n=99, we have:

1

------------------------ = 0.99..........9900.........0099.......99............

1.00...........001 100 9s 100 0s 100 9s

99 zeros

Consequently, there are precisely 100 nines followed by 100 zeros followed by 100 nines, and the pattern repeats indefinitely.

*Edited on ***June 20, 2022, 12:07 am**