Le Monde puzzle [#1133]

A weekly Monde current mathematical puzzle that reminded me of an earlier one (but was too lazy to check):

If ADULE-ELUDE=POINT, was is the largest possible value of POINT? With the convention that all letters correspond to different digits and no digit can start with 0. Same question when ADULE+ELUDE=POINT.

The run of a brute force R search return 65934 as the solution (codegolf welcomed!)

dify<-function(aluda,point) 
  (sum(aluda*10^(4:0))-sum(rev(aluda)*10^(4:0)))
num2dig<-function(dif) (dif%/%10^(0:4))%%10
sl=NULL
for (t in 1:1e6){
  adule=sample(0:9,5)
  while((dify(aluda)<=0)||(!prod(adule[c(1,5)])))
     adule=sample(0:9,5)
point=rev(num2dig(dify(adule)))
if ((!sum(duplicated(point)))&(prod(point%in%(0:9)[-adule-1])))
  sl=rbind(sl,c(adule,point))}
sl=as.matrix(distinct(as.data.frame(sl),.keep_all = TRUE))

where distinct is a dplyr R function.

> 94581-18549
[1] 76032

The code can be easily turned into solving the second question

> 31782+28713
[1] 60495

Le Monde puzzle [#1134]

A Monde mathematical puzzle on gcd’s and scm’s:

If one replaces a pair (a,b) of integers with the pair (g,s) of their greatest common denominator and smallest common multiple, how long at most before the sequence ends. Same question when considering a collection of five integers where two are selected by the pair (g,s) of their greatest common denominator and smallest common multiple.

The first question is straightforward as s is a multiple of s. So the sequence ends at most after one run. For five, run of a brute force R search return 9 as “the” solution (even though the true maximum is 10, as illustrated by the quintuplet (16,24,36,54,81):

ogcd <- function(x,y){r<-x%%y
  return(ifelse(r,ogcd(y,r),y))}

oscm<-function(x,y) x*y/ogcd(x,y)

divemul<-function(a,b) return(c(oscm(a,b),ogcd(a,b)))

for (t in 1:1e5){
ini=sample(1:1e2,5)
i=0;per=ker=sample(ini,2)
nez=divemul(per[1],per[2])
while(!max(nez%in%per)){
 ini=c(ini[!ini%in%per],nez)
 per=sample(ini,2)
 ker=rbind(ker,per)
 nez=divemul(per[1],per[2])
 i=i+1}
 sol=max(sol,i)}

LPPR or not?!

Le Monde puzzle [#1132]

A vaguely arithmetic challenge as Le weekly Monde current mathematical puzzle:

Given two boxes containing x and 2N+1-x balls respectively. If one proceeds by repeatedly transferring half the balls from the even box to the odd box, what is the largest value of N for which the resulting sequence in one of the boxes covers all integers from 1 to 2N?

The run of a brute force R search return 2 as the solution

lm<-function(N)
fils=rep(0,2*N)
bol=c(1,2*N)
while(max(fils)<2){
    fils[bol[1]]=fils[bol[1]]+1
    bol=bol+ifelse(rep(!bol[1]%%2,2),-bol[1],bol[2])*c(1,-1)/2}
return(min(fils))}

with obvious arguments that once the sequence starts cycling all possible numbers have been visited:

> lm(2)
[1] 1
> lm(3)
[1] 0

While I cannot guess the pattern, there seems to be much larger cases when lm(N) is equal to one, as for instance 173, 174, 173, 473, 774 (and plenty in-between).

Le Monde puzzle [#1129]

A number challenge as Le weekly Monde current mathematical puzzle:

When the three consecutive numbers 110, 111 and 112, they all are multiples of the sum of their digits. Are there 4 consecutive numbers with three digits like this? A contrario, does there exist 17 consecutive numbers with three digits such that they cannot be divided by the sum of their digits? 18?

The run of a brute force R search return 510,511,512,513 as the solution to the first question

library(gtools)
bez=!(100:999)%%apply(baseOf(100:999),1,sum)
> (100:897)[bez[-(1:3)]*bez[-c(1:2,900)]*bez[-c(1,899:900)]*bez[-(898:900)]==1]
[1] 510

And to the second one:

> max(diff((1:899)[!!diff(bez)]))
[1] 17

 

Le Monde puzzle [#1130]

A two-player game as Le weekly Monde current mathematical puzzle:

Abishag and Caleb fill in alternance a row of N boxes in a row by picking one then two then three &tc. consecutive boxes. When a player is unable to find enough consecutive boxes, the player has lost. Who is winning when N=29? When N=30?

Using a basic recursive search for the optimal strategy, with the status of the row and the number of required boxes as entries,

f<-function(b=!1:N,r=0){
  for(i in 1:(N-r)){
    if(p<-!max(b[j<-i+r:0])){
      q=b;q[j]=1
      if(p<-!f(q,r+1))break}}
  p}

returns Abishag as the winner for N=29 (as well as for N=1,2,7,…,13,19,…,29) and Caleb as the winner for N=30 (as well as for N=3,…,6,14,…,18). I am actually surprised that the recursion operates that deep, even though this means a √N depth for the recursion. While the code took too long to complete, the function operates for N=100. A side phenomenon is the apparent special case of N=47, which makes Abishag the looser, while N=46 and N=48 do not.This is an unusual pattern as otherwise (up to N=59), there are longer and longer stretches of adjacent wins and looses as N increases.

Le Monde puzzle [#1127]

A permutation challenge as Le weekly Monde current mathematical puzzle:

When considering all games between 20 teams, of which 3 games have not yet been played, wins bring 3 points, losses 0 points, and draws 1 point (each). If the sum of all points over all teams and all games is 516, was is the largest possible number of teams with no draw in every game they played?

The run of a brute force R simulation of 187 purely random games did not produce enough acceptable tables in a reasonable time. So I instead considered that a sum of 516 over 187 games means solving 3a+2b=516 and a+b=187, leading to 142 3’s to allocate and 45 1’s. Meaning for instance this realisation of an acceptable table of game results

games=matrix(1,20,20);diag(games)=0
while(sum(games*t(games))!=374){
  games=matrix(1,20,20);diag(games)=0
  games[sample((1:20^2)[games==1],3)]=0}
games=games*t(games)
games[lower.tri(games)&games]=games[lower.tri(games)&games]*
    sample(c(rep(1,45),rep(3,142)))* #1's and 3'
    (1-2*(runif(20*19/2-3)<.5)) #sign
games[upper.tri(games)]=-games[lower.tri(games)]
games[games==-3]=0;games=abs(games)

Running 10⁶ random realisations of such matrices with no constraint whatsoever provided a solution with] 915,524 tables with no no-draws, 81,851 tables with 19 teams with some draws, 2592 tables with 18 with some draws and 3 tables with 17 with some draws. However, given that 10*9=90 it seems to me that the maximum number should be 10 by allocating all 90 draw points to the same 10 teams, and 143 3’s at random in the remaining games, and I reran a simulated annealing version (what else?!), reaching a maximum 6 teams with no draws. Nowhere near 10, though!