## Le Monde puzzle [#1048]

Posted in Books, Kids, R with tags , , , , , on April 1, 2018 by xi'an

An arithmetic Le Monde mathematical puzzle:

A magical integer m is such that the remainder of the division of any prime number p by m is either a prime number or 1. What is the unique magical integer between 25 and 100? And is there any less than 25?

The question is dead easy to code

primz=c(1,generate_primes(2,1e6))
for (y in 25:10000)
if (min((primz[primz>y]%%y)%in%primz)==1) print(y)


and return m=30 as the only solution. Bon sang but of course!, since 30=2x3x5… (Actually, the result follows by dividing the quotient of the division of a prime number by 2 by 3 and then the resulting quotient by 5: all possible cases produce a remainder that is a prime number.) For the second question, the same code returns 2,3,4,6,8,12,18,24 as further solutions. There is no solution beyond 30.

## Le Monde puzzle [#967]

Posted in Books, Kids, pictures, Statistics, Travel, University life with tags , , , , , , , on September 30, 2016 by xi'an

A Sudoku-like Le Monde mathematical puzzle for a come-back (now that it competes with The Riddler!):

Does there exist a 3×3 grid with different and positive integer entries such that the sum of rows, columns, and both diagonals is a prime number? If there exist such grids, find the grid with the minimal sum?

I first downloaded the R package primes. Then I checked if by any chance a small bound on the entries was sufficient:

cale<-function(seqe){
ros=apply(seqe,1,sum)
cole=apply(seqe,2,sum)
dyag=sum(diag(seqe))
dayg=sum(diag(seqe[3:1,1:3]))
return(min(is_prime(c(ros,cole,dyag,dayg)))>0)}


Running the blind experiment

for (t in 1:1e6){
n=sample(9:1e2,1)
if (cale(matrix(sample(n,9),3))) print(n)}


I got 10 as the minimal value of n. Trying with n=9 did not give any positive case. Running another blind experiment checking for the minimal sum led to the result

> A
[,1] [,2] [,3]
[1,] 8 3 6
[2,] 1 5 7
[3,] 2 11 4


with sum 47.

## Solving the rectangle puzzle

Posted in Kids, R with tags , , , , , on March 16, 2010 by xi'an

Given the wrong solution provided in Le Monde and comments from readers, I went to look a bit further on the Web for generic solutions to the rectangle problem. The most satisfactory version I have found so far is Mendelsohn’s in Mathematics Magazine, which gives as the maximal number

$k^\star = (n+1)(n^2+n+1)$

for a $N\times N=(n^2+n+1)\times(n^2+n+1)$ grid. His theorem is based on the theory of projective planes and $n$ must be such that a projective plane of order $n$ exists, which seems equivalent to impose that $n$ is a prime number. The following graph plots the pairs $(N,k^\star)$ when $N=1,\ldots,13$ along with the known solutions, the fit being perfect for the values of $N$ of Mendelsohn’s form (i.e., 3, 7, 13).

Unfortunately, the formula does not extend to other values of $N$, despite Menselsohn’s comment that using for $n$ the positive root of the equation $x^2+x+1=N$ and then replacing $n$ by nearby integers (in the maximal number) should work. (The first occurrence I found of a solution for a square-free set did not provide a generic solution, but only algorithmic directions. While it is restricted to squares. the link with fractal theory is nonetheless interesting.)