Answer Posted / y@$w@nth

s9(9)v9(9)

Here v stands for assumed decimal point so it will occupy
only 18 bytes(9+9)...if instead of s9(9)v9(9) it is coded as
s9(9).9(9) then it is going to occupy 19 bytes ((9+9)+1)
the extra 1 byte for decimal point .........There is no way
of occupying 20 bytes storage.

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