Starting at the decimal point ...
-- Look at all the digits in the first place after the decimal point.
Find the smallest one. If more than one number has that digit
in the first place, then keep those, discard the others, and ...
-- Look at the digits in the second place after the decimal point.
Find the smallest one. If more than one number has that digit
in the second place, then keep those, discard the others, and ...
-- Look at the digits in the third place after the decimal point.
Find the smallest one. If more than one number has that digit
in the third place, then keep those, discard the others, and ...
-- Look at all the digits in the fourth place after the decimal point.
Find the smallest one. If more than one number has that digit in
the fourth place, then keep those, discard the others and ...
.
.
.
etc.
Keep going until you have only one number left. That's the smallest one,
out of the entire original list.
In the example you gave, the smallest digit in the first place after the decimal point
is the '0', and choice-'C' is the only one that has it. So choice-'C' is the least, and
you don't need to go any farther.
Here's another one to practice on. It needs the whole procedure that I wrote out,
but now it should be easy for you:
0.1234607
0.1234915
0.1234095
0.1234017
0.1234184
0.1234521
