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
Puzzle Solution
