## Le Monde puzzle [#1139]

Posted in Books, Kids, R, Statistics with tags , , , , , on April 24, 2020 by xi'an A weekly Monde current mathematical puzzle that reminded me of an earlier one (but was too lazy to check):

The integer n=36 enjoys the property that all the differences between its ordered divisors are also divisors of 36. Find the only 18≤m≤100 that enjoys this property such that all its prime dividers areof multiplicity one. Are there other such m’s?

The run of a brute force R search return 42 as the solution (codegolf welcomed!)

y=z=1:1e5
for(x in y)z[y==x]=!sum(x%%diff((1:x)[!x%%(1:x)]))
y=y[z==1]
for(k in generate_primes(2,max(y)))y=y[!!y%%k^2]


where generate_primes is a primes R function. Increasing the range of y’s to 10⁵ exhibits one further solution, 1806.

## Le Monde puzzle [#1124]

Posted in Books, Kids, R with tags , , , , , on December 29, 2019 by xi'an A prime number challenge [or rather two!] as Le weekly Monde current mathematical puzzle:

When considering the first two integers, 1 and 2, their sum is 3, a prime number. For the first four integers, 1,2,3,4, it is again possible to sum them pairwise to obtain two prime numbers, eg 3 and 7. Up to which limit is this operation feasible? And how many primes below 30,000 can write as n^p+p^n?

The run of a brute force R simulation like

max(apply(apply(b<-replicate(1e6,(1:n)+sample(n)),2,is_prime)[,b[1,]>2],2,prod))


provides a solution for the first question until n=14 when it stops. A direct explanation is that the number of prime numbers grows too slowly for all sums to be prime. And the second question gets solved by direct enumeration using again the is_prime R function from the primes package:

 1 1
 1 2
 1 4
 2 3
 3 4


## Le Monde puzzle [#1048]

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

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.)