**T**his may sound like an absurd question [and in some sense it is!], but this came out of a recent mathematical riddle on The Riddler, asking for the largest number one could write with ten symbols. The difficulty with this riddle is the definition of a symbol, as the collection of available symbols is a very relative concept. For instance, if one takes the symbols available on a basic pocket calculator, besides the 10 digits and the decimal point, there should be the four basic operations plus square root and square,which means that presumably 999999999² is the largest one can on a cell phone, there are already many more operations, for instance my phone includes the factorial operator and hence 9!!!!!!!!! is a good guess. While moving to a computer the problem becomes somewhat meaningless, both because there are very few software that handle infinite precision computing and hence very large numbers are not achievable without additional coding, and because it very much depends on the environment and on which numbers and symbols are already coded in the local language. As illustrated by this X validated answer, this link from The Riddler, and the xkcd entry below. (The solution provided by The Riddler itself is not particularly relevant as it relies on a particular collection of symbols, which mean Rado’s number BB(9999!) is also a solution within the right referential.)

## Archive for factorial

## how large is 9!!!!!!!!!?

Posted in Statistics with tags big numbers, cellphone, coding, cross validated, factorial, mathematical puzzle, pocket calculator, Rado's numbers, The Riddler, Turing machine on March 17, 2017 by xi'an## Puzzle of the week [26]

Posted in Statistics with tags factorial, Le Monde, mathematical puzzle, number theory on July 3, 2010 by xi'an* Le Monde* weekend puzzle for the past week (I have not peeked at the solution yet!) was quite straightforward: find those

*n*‘s such that

*2*divides

^{n}*n!*and those

*n*‘s for which

*2*divides

^{n-1}*n*. Looking at the problem in the plane to Montpellier, I think that the solution is that no positive

*n*exists such that

*2*divides n! and powers of 2 are those numbers for which

^{n}*2*divides

^{n-1}*n*.

**M**y reasoning is

- that the numbers with the highest potential is a power of 2,
- that
*2*divides n! when^{n-1}*n*is of the form*2*, and^{m} - therefore that no integer n can satisfy the harder constraint.

Proving that *2 ^{n-1}* divides

*n!*when

*n*is of the form

*n=2*can be done by induction: it works for

^{m}*n=2*and if it works for

*2*, then it works for

^{m}*n=2*by considering a separation of

^{m+1}*n!*into

and by using the induction assumption that *(2 ^{m})!* can be divided by

Recycling the dividers for the second part leads to its being divisible by

because the very last term in the factorial is by *2 ^{m+1}*, which can be divided by

*2*… Proving that integers other than the

^{m+1}*2*‘s cannot be divided by

^{m}*2*again works by an induction proof on

^{n-1 }*m*.