Archive for mathematical puzzle

Le Monde puzzle [#977]

Posted in Books, Kids, pictures, Statistics, Travel, University life with tags , , , , , on October 3, 2016 by xi'an

A mild arithmetic Le Monde mathematical puzzle:

Find the optimal permutation of {1,2,..,15} towards minimising the maximum of sum of all three  consecutive numbers, including the sums of the 14th, 15th, and first numbers, as well as the 15th, 1st and 2nd numbers.

If once again opted for a lazy solution, not even considering simulated annealing!,


and got the minimal value of 26 for the permutation

14 9 2 15 7 1 11 10 4 12 8 5 13 6 3

Le Monde gave a solution with value 25, though, which is

11 9 7 5 13 8 2 10 14 6 1 12 15 4 3

but there is a genuine mistake in the solution!! This anyway shows that brute force may sometimes fail. (Which sounds like a positive conclusion to failing to find the proper solution! But trying with a simple simulated annealing version did not produce any 25 either…)

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:


Running the blind experiment

for (t in 1:1e6){
  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.

Le Monde puzzle [#965]

Posted in Kids, R with tags , , , on June 14, 2016 by xi'an

A game-related Le Monde mathematical puzzle:

Starting with a pile of 10⁴ tokens, Bob plays the following game: at each round, he picks one of the existing piles with at least 3 tokens, takes away one of the tokens in this pile, and separates the remaining ones into two non-empty piles of arbitrary size. Bob stops when all piles have identical size. What is this size and what is the maximal number of piles?

First, Bob can easily reach a decomposition that prevents all piles to be of the same size: for instance, he can start with a pile of 1 and another pile of 2. Looking at the general perspective, an odd number of tokens, n=2k+1, can be partitioned into (1,1,2k-1). Which means that the decomposition (1,1,…,1) involving k+1 ones can always be achieved. For an even number, n=2k, this is not feasible. If the number 2k can be partitioned into equal numbers u, this means that the sequence 2k-(u+1),2k-2(u+1),… ends up with u, hence that there exist m such that 2k-m(u+1)=u or that 2k+1 is a multiple of (u+1). Therefore, the smallest value is made of the smallest factor of 2k+1. Minus one. For 2k=10⁴, this value is equal to 72, while it is 7 for 10³. The decomposition is impossible for 2k=100, since 101 is prime. Here are the R functions used to check this analysis (with small integers, if not 10⁴):

solvant <- function(piles){
 if ((length(piles)>1)&((max(piles)==2)||(min(piles)==max(piles)))){
   while (piles[i]<3)

disolvant <- function(piles){
 while (min(sol)<max(sol))

resolvant <- function(piles){
 for (t in 1:piles){
 if (length(sol)>maxle){

Le Monde puzzle [#964]

Posted in Books, Kids, R with tags , , on June 2, 2016 by xi'an

A not so enticing Le Monde mathematical puzzle:

Find the minimal value of a five digit number divided by the sum of its digits.

This can formalised as finding the minimum of N/(a+b+c+d+e) when N writes abcde. And solved by brute force. Using a rough approach to finding the digits of a five-digit number, the question can be easily solved as

for (i in 1e4:1e5){
 pres=i/sum((i %% 10^{5:1})) %/% 10^{4:0})
 if (pres<pris){

which returns N=10999 as its solution. (The solution for six digits is 10999.) The mathematical solution as provided in the newspaper certainly sounded more exciting.

a Simpson paradox of sorts

Posted in Books, Kids, pictures, R with tags , , , , , , , , , on May 6, 2016 by xi'an

The riddle from The Riddler this week is about finding an undirected graph with N nodes and no isolated node such that the number of nodes with more connections than the average of their neighbours is maximal. A representation of a connected graph is through a matrix X of zeros and ones, on which one can spot the nodes satisfying the above condition as the positive entries of the vector (X1)^2-(X^21), if 1 denotes the vector of ones. I thus wrote an R code aiming at optimising this target

targe <- function(F){

by mere simulated annealing:

rate <- function(N){ 
# generate matrix F
# 1. no single 
for (i in 2:(N-1)){ 
if (sum(F[,i])==0) 
if (sum(F[,N])==0) 
# 2. more connections 
#simulated annealing
for (t in 1:T){
  #1. local proposal
  while (min(prop%*%rep(1,N))==0){
  if (log(runif(1))*temp<target-targo){ 
#2. global proposal 
  prop=F prop[lower.tri(prop)]=F[lower.tri(prop)]+
  if (log(runif(1))*temp<target-targo){

Eward SimpsonThis code returns quite consistently (modulo the simulated annealing uncertainty, which grows with N) the answer N-2 as the number of entries above average! Which is rather surprising in a Simpson-like manner since all entries but two are above average. (Incidentally, I found out that Edward Simpson recently wrote a paper in Significance about the Simpson-Yule paradox and him being a member of the Bletchley Park Enigma team. I must have missed out the connection with the Simpson paradox when reading the paper in the first place…)

Le Monde puzzle [#960]

Posted in Kids, R with tags , , , , , on April 28, 2016 by xi'an

An arithmetic Le Monde mathematical puzzle:

Given an integer k>1, consider the sequence defined by F(1)=1+1 mod k, F²(1)=F(1)+2 mod k, F³(1)=F²(1)+3 mod k, &tc. [With this notation, F is not necessarily a function.] For which value of k is the sequence the entire {0,1,…,k-1} set?

This leads to an easy brute force resolution, for instance writing the R function

crkl<-function(k) return(unique(cumsum(1:(2*k))%%k))

where 2k is a sufficient substitute for ∞. Then the cases where the successive images of 1 visit the entire set {0,1,…,k-1} are given by

> for (i in 2:550) if (length(crkl(i))==i) print(i)
[1] 2
[1] 4
[1] 8
[1] 16
[1] 32
[1] 64
[1] 128
[1] 256
[1] 512

which suspiciously looks like the result that only the powers of 2 k=2,2²,2³,… lead to a complete exploration of the set {0,1,…,k-1}. Checking a few series in the plane back from Warwick, I quickly found that when k is odd, (1) the sequence is of period k and (2) there is symmetry in the sequence, which means it only takes (k-1)/2 values. For k even, there is a more complicated symmetry, with the sequence being of period 2k, symmetric around its two middle values, and taking the values 1,2,..,1+k(2k+1)/4,..,1+k(k+1)/2. Those values cannot cover the set {0,1,…,k-1} if two are equal, which means an i(i+1)/2 congruent to zero modulo k, hence equal to k. This is clearly impossible when k is a power of 2 because i and i+1 cannot both divide a power of 2. I waited for the published solution as of yesterday’s and the complete argument was to show that when N=2p, the corresponding sequence [for N] is made (modulo p) of the sequence for p plus the same sequence translated by p. The one for N is complete only if the one for p is complete, which by recursion eliminates all cases but the powers of 2…

Le Monde puzzle [#959]

Posted in Kids, R with tags , , , on April 20, 2016 by xi'an

Another of those arithmetic Le Monde mathematical puzzle:

Find an integer A such that A is the sum of the squares of its four smallest dividers (including1) and an integer B such that B is the sum of the third poser of its four smallest factors. Are there such integers for higher powers?

This begs for a brute force resolution checking the integers until a solution appears. The only exciting part is providing the four smallest factors but a search on Stack overflow led to an existing R function:

FUN <- function(x) {
    x <- as.integer(x)
    div <- seq_len(abs(x))
    return(div[x %% div == 0L])

(which uses the 0L representation I was unaware of) and hence my R code:

 while ((stop)&(I<1e6)){
  if (length(dive)>3)

But this code only seems to work for n=2 as produces A=130: it does not return any solution for the next value of n… As shown by the picture below, which solely exhibits a solution for n=2,5, A=17864 (in the second case), there is no solution less than 10⁶ for n=3,4,6,..9. So, unless I missed a point in the question, the solutions for n>2 are larger if they at all exist.

lemonde959A resolution got published yesterday night in Le Monde  and (i) there is indeed no solution for n=3 (!), (ii) there are solutions for n=4 (1,419,874) and n=5 (1,015,690), which are larger than the 10⁶ bound I used in the R code, (iii) there is supposedly no solution for n=5!, when the R code found that 17,864=1⁵+2⁵+4⁵+7⁵… It is far from the first time the solution is wrong or incomplete!