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## Sagot :

-- Look at all the digits in the

**place after the decimal point.**

__first__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

**place after the decimal point.**

__second__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

**place after the decimal point.**

__third__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

**place after the decimal point.**

__fourth__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