## Le Monde puzzle [#1028]

Posted in Books, Kids with tags , , , on November 16, 2017 by xi'an Back to standard Le Monde mathematical puzzles (no further competition!), with this arithmetic one:

While n! cannot be a squared integer for n>1, does there exist 1<n<28 such that 28(n!) is a square integer? Does there exist 1<n,m<28 such that 28(n!)(m!) is a square integer? And what is the largest group of distinct integers between 2 and 27 such that the product of 28! by their factorials is a square?

The fact that n! cannot be a square follows from the occurrence of single prime numbers in the resulting prime number decomposition. When considering 28!, there are several single prime numbers like 17, 19, and 23, which means n is at least 23, but then the last prime in the decomposition of 28! being 7 means this prime remains alone in a product by any n! when n<28. However, to keep up with the R resolution tradition, I started by representing all integers between 2 and 28 in terms of their prime decomposition:

primz=c(2,3,5,7,11,13,17,19,23)
dcmpz=matrix(0,28,9)
for (i in 2:28){
for (j in 1:9){
k=i
while (k%%primz[j]==0){
k=k%/%primz[j];dcmpz[i,j]=dcmpz[i,j]+1}}
}


since the prime number factorisation of the factorials n! follows by cumulated sums (over the rows) of dcmpz, after which checking for one term products

fctorz=apply(dcmpz,2,cumsum)
for (i in 23:28)
if (max((fctorz[28,]+fctorz[i,])%%2)==0) print(i)


and two term products

for (i in 2:28)
for (j in i:27)
if (max((fctorz[28,]+fctorz[i,]+fctorz[j,])%%2)==0)
print(c(i,j))

is easy and produces i=28 [no solution!] in the first case and (i,j)=(10,27) in the second case. For the final question,  adding up to twelve terms together still produced solutions so I opted for the opposite end by removing one term at a time and

for (a in 2:28)
if (max(apply(fctorz[-a,],2,sum)%%2)==0) print(a)


exhibited a solution for a=14. Meaning that

2! 3! …. 13! 15! …. 28!

is a square.

## how large is 9!!!!!!!!!?

Posted in Statistics with tags , , , , , , , , , on March 17, 2017 by xi'an This 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.) ## Puzzle of the week 

Posted in Statistics with tags , , , 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 2n divides n! and those n‘s for which 2n-1 divides n. Looking at the problem in the plane to Montpellier, I think that the solution is that no positive n exists such that 2n divides n! and powers of 2 are those numbers for which 2n-1 divides n.

My reasoning is

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

Proving that 2n-1 divides n! when n is of the form n=2m can be done by induction: it works for n=2 and if it works for 2m, then it works for n=2m+1 by considering a separation of n! into $(2^m)! \times (2^m+1)\cdots 2^{m+1}$

and by using the induction assumption that (2m)! can be divided by $2^{2^m-1}.$

Recycling the dividers for the second part leads to its being divisible by $2^{2^m}$

because the very last term in the factorial is by 2m+1, which can be divided by 2m+1… Proving that integers other than the 2m‘s cannot be divided by 2n-1 again works by an induction proof on m.