All numbers between 1 and 1010 are written out and ordered alphabetically into a dictionary (as the only entries).
Spaces and hyphens are removed. 1024 would then be "onethousandtwentyfour".
Also, "and" is not being used as a link (I think that's American English).
Furthermore, the American number scale is used: million, billion, trillion...
Myriad is not used, and numbers with (or like) 1984 are composed with "onethousand",
so "nineteeneightyfour" would not be allowed.
Last but not least: the numbers are written out in, well, (American) English.
What is the first odd number?
Bonus question: What is the first prime number?
Answer
Number words in the given range are assembled from the following components:
- The set $A$ of single digits:
eight
,five
,four
,nine
,one
,seven
,six
,three
,two
. - The set $B$ of double-digit blocks:
eighteen
,eleven
,fifteen
,fourteen
,nineteen
,seventeen
,sixteen
,thirteen
,twelve
. - The set $C$ of tens digits:
eighty
,fifty
,forty
,ninety
,seventy
,sixty
,thirty
,twenty
. hundred
- The powers of 1000:
billion
,million
,thousand
.
Number words are generated by the following non-ambiguous grammar (I use the abbreviated notation $X^?$ to mean $X$ or nothing): $$ \begin{array}{lllll} S \to & & & T \, \texttt{thousand} & T^? \\ S \to & & T \, \texttt{million} & & T^? \\ S \to & & T \, \texttt{million} & T \, \texttt{thousand} & T^? \\ S \to & A \, \texttt{billion} & & & T^? \\ S \to & A \, \texttt{billion} & & T \, \texttt{thousand} & T^? \\ S \to & A \, \texttt{billion} & T \, \texttt{million} & & T^? \\ S \to & A \, \texttt{billion} & T \, \texttt{million} & T \, \texttt{thousand} & T^? \\ T \to & & A \\ T \to & & B \\ T \to & & C \, A^? \\ T \to & A \, \texttt{hundred} & B \\ T \to & A \, \texttt{hundred} & C^? \, A^? \\ \end{array} $$
The constraint that the number is odd implies that the rightmost non-terminal will be $A$ or $B$ (otherwise the number is a multiple of $10$). Every production that has $A$ or $B$ at the rightmost position allows the rightmost non-terminals to be $A$, $B$, or $C \, A$. The first odd $A$ is five
, the first odd $B$ is eleven
, and the first odd $C \, A$ is eightyfive
. The winner is eightyfive
: that's the first odd number in the range $[1,99]$, and that's how the first odd number ends.
Going left, eighthundred
competes with leaving out the $A \, \texttt{hundred}$ part. Since eighthundred
comes before eightyfive
, the first number ends with eighthundredeightyfive
(which is the first odd number in the range $[1,999]$).
For the other occurrences of $T$ (in front of thousand
and million
), we don't have the constraint that the number must be odd. There eight
followed by a power of 1000 competes with eighteen
(which beats eighthundred…
); for both thousand
and million
, eighteen
wins. If billion
appears, then the first member of $A$ (eight
) is in front. Thus the following numbers compete:
eighthundredeightyfive
eighteenthousand eighthundredeightyfive
eighteenmillion eighthundredeightyfive
eighteenmillion eighteenthousand eighthundredeightyfive
eightbillion eighthundredeightyfive
eightbillion eighteenthousand eighthundredeightyfive
eightbillion eighteenmillion eighthundredeightyfive
eightbillion eighteenmillion eighteenthousand eighthundredeightyfive
eightbillion
wins against eighteen…
. As for the powers of 1000 to include prefixed by eighteen
when they all start after een
, pick the first one in alphabetical order, then the first one that's numerically smaller, and so on. Or, with only four possibilities, you can check manually.
The winner is:
eightbillion eighteenmillion eighteenthousand eighthundredeightyfive
or 8,018,018,885 for short. This would still be the first word for an odd number if the dictionary included all number words with conventionally-accepted names (in U.S. English, without and
).
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