When is the next "Factorion" date?

For reference, here's a table of all the usable factorials:

where 7! and beyond are too large to fit in any nearby year until 5040. Second, note that the bulk of the sum will be from 5!s and 6!s, and we can't easily fit those in some of the digits: The tens places of the month and day are limited, and we want to avoid making the hundreds and thousands places large. So, realistically, the bulk of our sum needs to be composed of 4 (or maybe 5 if we can use the hundreds place) factorials of 5 or 6.

We have $6! \cdot 3 = 720 \cdot 3 = 2160$, so if we want to make any year between now and $2160$, we need to use at most 2 copies of $6!$ - but that leaves us with two copies of $5!$ for our four unbounded digits, making $2 \cdot 5! + 2 \cdot 6! = 1680$, nowhere near enough to make any recent year. So let's look at and around 2160.

Note that any year between 2167 and 2199 would contain a digit between 7 and 9, making the sum far too large, so here we're trying to make some year between 2160 and 2166. Any of these years starts with 21, so the sum of digits for any usable date around these is going to be at least $2! + 1! + 6! \cdot 3 = 2163$, and using 2164 or 2165 would result in too large a sum. So, we're trying to find a usable date in 2166.

Using the MM and DD digits, we need to create $2166 - (2! + 1! + 6! + 6!) = 2166 - 1443 = 723$ - one copy of $6!$, plus three - which we can make with three copies of $0!$ or $1!$. To make this as early as possible in 2166, we want the month to be as small as possible - so make the month 01, January, and then the date is 06.

So the next factorion date is $06/01/2166$, or the 6th of January 2166.

As for the other questions, there are clearly finitely many factorion dates, since 999...999 will eventually outgrow 9! + 9! + 9! + ... + 9!. To find the most recent previous one:

• By a similar argument to before, we can't build a large enough sum to make any year from 2000-2025.
• Any year from 1667-1999 contains a digit too large to be usable
• 1666 results in a too large sum on its own
• Two copies of $6!$ and two copies of $5!$ make $1680$, outside the usable range, so our date will have two $6$s and at most one $5$.
• This leaves us with a sum that's going to be at most around $6! + 6! + 5! + 4! + 4! = 1608$, but anything near here is going to have a tens digit too low to be used as a large digit, so we can't achieve anything past $1600$
• Like before, anything from $1567$ to $1600$ has a 7, 8 or 9, so all of these are unusable.

So let's try $1566$. With the remaining four digits, we need to make up a sum of $1566 - 1! - 5! - 6! - 6! = 5$, which we can make with three $0!$s or $1!$s and one $2!$. Since we're trying to make the date as late as possible, we want to make the month as large as possible, and we can make it be 12 since we have a 2. Then, the largest we can make the day is 11.

So the most recent previous factorion date is $11/12/1566$, or the 11th of December 1566.

Finally, let's try to make the last factorion date. (this was ... difficult) The remaining digit factorials are $7! = 5040$, $8! = 40320$, and $9! = 362880$. We want to make the sum as large as possible, so ideally we're looking at a 7-digit year of which we can feasibly make 5-7 digits $9$s. The relevant multiples of $9!$ are

• $9! \cdot 7 = 2540160$ - which would need to be $x9/09/2599999$, which can't be feasibly made
• $9! \cdot 6 = 2177280$ - which would need to have at least four of the last 5 digits be 9s
• $9! \cdot 5 = 1814400$ - which would need to have at least three of the last 5 digits be 9s

(at this point the numbers were getting annoying so I started using a calculator, but you could do this by hand)

So let's look at $9! \cdot 6 = 2177280$. Note that, ignoring the necessary $9$s, the largest amount we can increase this by is $8! = 40320$, so anything beyond 2200000 won't give us enough large digits (6 9s and at least one 8) to make things work. Next, note that the 7 and 8 in 2179999 and 2189999 would push the sum too far, and we would need at least three 7s to get to 219xxxx. So we can't have six 9s.

Moving on, let's try and make something around $9! \cdot 5 = 1814400$. This already has an $8$ in it, so we're looking around $1814400 + 8! = 1850720$, where at least three of the trailing digits need to be 9s.

For the sake of making this as large as possible, let's try to make 189xxxx by adding another 8. Then we have the sum at $1850720 + 40320 = 1891040$, and the remaining digits need to contain another 8 and four more 9s. But to get the hundreds and thousands digits high enough to be usable would require more 6s and 7s than we can spare, so that won't work.

Similarly, to get something around 1850720 to work, we need to turn at least three digits to 9s using as few factorials as possible. The limitations of adding factorials really start to become apparent at this point - 1850999 is the only thing in this range that could realistically work, and even that falls victim to the digits not being flexible enough. So, five 9s won't work, and any solution can have at most 4.

$4 \cdot 9! = 1451520$, and at least two of the trailing digits need to be 9s. At this point, the search space is quite large and it's difficult to rule out other options, but we can try making 149xxxx by adding a hypothetical 8! to get 1491840. Then, with a hypothetical 5!, we can turn this into 1491960, which gets two 9s. Adding the two 1!s and 4! to our total gives $(\text{extra} 5! + 9! + 9!) + 1! + 4! + 9! + 1! + 9! + 8! = 1491986$ - and by adding 3, in the form of $2! + 1$, we can get 1491989, needing an extra 9, 5, 2 and 0 or 1 in the extra digits. The 5 and 9 need to be the last digits, so we can have either 29/05 or 25/09, of which the latter is later.

So, the final factorion date is probably 25th September, 1491989.