Archive for mathematical puzzle

Le Monde puzzle [#1085]

Posted in Books, Kids, R with tags , , , , , on February 18, 2019 by xi'an

A new Le Monde mathematical puzzle in the digit category:

Given 13 arbitrary relative integers chosen by Bo, Abigail can select any subset of them to be drifted by plus or minus one by Bo, repeatedly until Abigail reaches the largest possible number N of multiples of 5. What is the minimal possible value of N under the assumption that Bo tries to minimise it?

I got stuck on that one, as building a recursive functiion led me nowhere: the potential for infinite loop (add one, subtract one, add one, …) rather than memory issues forced me into a finite horizon for the R function, which then did not return anything substantial in a manageable time. Over the week and the swimming sessions, I thought of simplifying the steps, like (a) work modulo 5, (b) bias moves towards 1 or 4, away from 2 and 3, by keeping only one entry in 2 and 3, and all but one at 1 and 4, but could only produce five 0’s upon a sequence of attempts… With the intuition that only 3 entries should remain in the end, which was comforted by Le Monde solution the week after.

Le Monde puzzle [#1083]

Posted in Books, Kids, R, Travel with tags , , , , , , on February 7, 2019 by xi'an

A Le Monde mathematical puzzle that seems hard to solve without the backup of a computer (and just simple enough to code on a flight to Montpellier):

Given the number N=2,019, find a decomposition of N as a sum of non-trivial powers of integers such that (a) the number of integers in the sum is maximal or (b) all powers are equal to 4.  Is it possible to write N as a sum of two powers?

It is straightforward to identify all possible terms in these sums by listing all powers of integers less than N

for (pow in 3:11)

which leads to 57 distinct powers. Sampling at random from this collection at random produces a sum of 21 perfect powers:


But looking at the 22 smallest numbers in the pool of powers leads to 2019, which is a sure answer. Restricting the terms to powers of 4 leads to the sequence

1⁴+2⁴+3⁴+5⁴+6⁴ = 2019

And starting from the pools of all possible powers in a decomposition of 2019 as the sum of two powers shows this is impossible.

missing digit in a 114 digit number [a Riddler’s riddle]

Posted in R, Running, Statistics with tags , , , , , , , on January 31, 2019 by xi'an

A puzzling riddle from The Riddler (as Le Monde had a painful geometry riddle this week): this number with 114 digits


is missing one digit and is a product of some of the integers between 2 and 99. By comparison, 76! and 77! have 112 and 114 digits, respectively. While 99! has 156 digits. Using WolframAlpha on-line prime factor decomposition code, I found that only 6 is a possible solution, as any other integer between 0 and 9 included a large prime number in its prime decomposition:

However, I thought anew about it when swimming the next early morning [my current substitute to morning runs] and reasoned that it was not necessary to call a formal calculator as it is reasonably easy to check that this humongous number has to be divisible by 9=3×3 (for else there are not enough terms left to reach 114 digits, checked by lfactorial()… More precisely, 3³³x33! has 53 digits and 99!/3³³x33! 104 digits, less than 114), which means the sum of all digits is divisible by 9, which leads to 6 as the unique solution.


Le Monde puzzle [#1081]

Posted in Books, Kids, R, Travel with tags , , , , on January 24, 2019 by xi'an

A “he said-she said” Le Monde mathematical puzzle (again in the spirit of the famous Singapore high-school birthdate problem):

Abigail and Corentin are both given a positive integer, a and b, such that a+b is either 19 or 20. They are asked one after the other and repeatedly if they are sure of the other’s number. What is the maximum number of times they are questioned?

If Abigail is given a 19, b=1 necessarily. Hence if Abigail does not reply, a<19. This implies that, if Corentin is given b=1 or b=19, he can reply a+b=19 or a+b=20, necessarily. Else, 1<b<19 implies that, if a=1 or a=18, b=18 or b=2. And so on…which leads to a maximum of 20 questions, 10 for Abigail and 10 for Corentin. Here is my R implementation

while ((max(az)>0)&(max(bz)>0)){
 if (at){ 
  for (i in 1:19){ 
   if (sum(az[i,]>0)==2){
   for (j in az[i,az[i,]>0]){ 
     if (sum(bz[j,]==0)==2) az[i,]=rep(0,2)}}
   if (sum(az[i,]>0)<2){ 
  if (bt){ 
   for (i in 1:19){ 
    if (sum(bz[i,bz[i,]>0]>0)==2){
     for (j in bz[i,bz[i,]>0]){ 
      if (sum(az[j,]==0)==2) bz[i,]=rep(0,2)}}
     if (sum(bz[i,]>0)<2){ bz[i,]=rep(0,2)}}}

Le Monde puzzle [#1076]

Posted in Books, Kids, R, Travel with tags , , , , , , , , , on December 27, 2018 by xi'an

A cheezy Le Monde mathematical puzzle : (which took me much longer to find [in the sense of locating] than to solve, as Warwick U does not get a daily delivery of the newspaper [and this is pre-Brexit!]):

Take a round pizza (or a wheel of Gruyère) cut into seven identical slices and turn one slice upside down. If the only possibly moves are to turn three connected slices to their reverse side, how many moves at least are needed to recover the original configuration? What is the starting configuration that requires the largest number of moves?

Since there are ony N=2⁷ possible configurations, a brute force exploration is achievable, starting from the perfect configuration requiring zero move and adding all configurations found by one additional move at a time… Until all configurations have been visited and all associated numbers of steps are stable. Here is my R implementation

nztr=lengz=rep(-1,N) #length & ancestor
fundz=matrix(0,Z,Z) #Z=7
for (i in 1:Z){ #only possible moves
while (min(lengz)==-1){ #second loop omitted
  for (j in (1:N)[lengz>-1])
  for (k in 1:Z){
    if ((lengz[m]==-1)|(lengz[m]>lengz[j]+1)){

Which produces a path of length five returning (1,0,0,0,0,0,0) to the original state:

> nztry(2)
[1] 1 0 0 0 0 0 0
[1] 0 1 1 0 0 0 0
[1] 0 1 0 1 1 0 0
[1] 0 1 0 0 0 1 0
[1] 1 1 0 0 0 0 1
[1] 0 0 0 0 0 0 0

and a path of length seven in the worst case:

> nztry(2^7)
[1] 1 1 1 1 1 1 1
[1] 1 1 1 1 0 0 0
[1] 1 0 0 0 0 0 0
[1] 0 1 1 0 0 0 0
[1] 0 1 0 1 1 0 0
[1] 0 1 0 0 0 1 0
[1] 1 1 0 0 0 0 1
[1] 0 0 0 0 0 0 0

Since the R code was written for an arbitrary number Z of slices, I checked that there is no solution for Z being a multiple of 3.

Le Monde puzzle [#1075]

Posted in Books, Kids, R with tags , , , , on December 12, 2018 by xi'an

A new Le Monde mathematical puzzle in the digit category:

Find the largest number such that each of its internal digits is strictly less than the average of its two neighbours. Same question when all digits differ.

For instance, n=96433469 is such a number. When trying pure brute force (with the usual integer2digits function!)

while (length(solz)>0){
 for (i in (10^(le+1)-1):(9*10^le+9)){
  x=as.numeric(strsplit(as.character(i), "")[[1]])
 if (min(x[-c(1,le+1)]<(x[-c(1,2)]+x[-c(le,le+1)])/2)==1){ print(i);solz=c(solz,i); break()}}

this is actually the largest number returned by the R code. There is no solution with 9 digits. Adding an extra condition

while (length(solz)>0){
 for (i in (10^(le+1)-1):(9*10^le+9)){
  x=as.numeric(strsplit(as.character(i), "")[[1]])
 if ((min(x[-c(1,le+1)]<(x[-c(1,2)]+x[-c(le,le+1)])/2)==1)&
    (length(unique(x))==le+1)){ print(i);solz=c(solz,i); break()}}

produces n=9520148 (seven digits) as the largest possible integer.

Le Monde puzzle [#1078]

Posted in Books, Kids, R with tags , , , , , , on November 29, 2018 by xi'an

Recalling Le Monde mathematical puzzle  first competition problem

Given yay/nay answers to the three following questions about the integer 13≤n≤1300 (i) is the integer n less than 500? (ii) is n a perfect square? (iii) is n a perfect cube?  n cannot be determined, but it is certain that any answer to the fourth question (iv) are all digits of n distinct? allows to identify n. What is n if the answer provided for (ii) was false.

When looking at perfect squares less than 1300 (33) and perfect cubes less than 1300 (8), there exists one single common integer less than 500 (64) and one single above (729). Hence, it is not possible that answers to (ii) and (iii) are both positive, since the final (iv) would then be unnecessary. If the answer to (ii) is negative and the answer to (iii) is positive, it would mean that the value of n is either 512 or 10³ depending on the answer to (i), excluding numbers below 500 since there is no unicity even after (iv). When switching to a positive answer to (ii), this produces 729 as the puzzle solution.

Incidentally, while Amic, Robin, and I finished among the 25 ex-aequos of the competition, none of us reached the subsidiary maximal number of points to become the overall winner. It may be that I will attend the reward ceremony at Musée des Arts et Métiers next Sunday.