Le Monde puzzle [#1067]

The second Le Monde mathematical puzzle in the new competition is sheer trigonometry:

When in the above figures both triangles ABC are isosceles and the brown segments are all of length 25cm, find the angle in A and the value of DC², respectively.

This could have been solved by R coding the various possible angles of the three segments beyond BC until the isosceles property is met, but it went much faster by pen and paper, leading to an angle of π/9 in the first case and a square of 1250 in the second case. The third puzzle is basic arithmetic that only seems solvable by enumeration…

Le Monde puzzle [#1066]

Recalling Le Monde mathematical puzzle  first competition problem

For the X table below, what are the minimal number of lights that are on (green) to reach the minimal and maximal possible numbers of entries (P) with an even (P as pair) number of neighbours with lights on? In the illustration below, there are 16 lights on (green) and 31 entries with an even number of on-neighbours.

As suggested last week, this was amenable to a R resolution by low-tech simulated annealing although the number of configurations was not that large when accounting for symmetries. The R code is a wee bit long for full reproduction here but it works on moving away from a random filling of this cross by 0’s and 1’s, toward minimising or maximising the number of P’s, this simulated annealing loop being inserted in another loop recording the minimal number of 1’s in both cases. A first round produced 1 and 44 for the minimal and maximal numbers of P’s, respectively, associated with at least 16 and 3 1’s, respectively, but a longer run exhibited 45 for 6 1’s crossing one of the diagonals of the X, made of two aligned diagonals of the outer 3×3 tables. (This was confirmed by both Amic and Robin in Dauphine!) The next [trigonometry] puzzle is on!

Le Monde puzzle [#1650]

A penultimate Le Monde mathematical puzzle  before the new competition starts [again!]

For a game opposing 40 players over 12 questions, anyone answering correctly a question gets as reward the number of people who failed to answer. Alice is the single winner: what is her minimal score? In another round, Bob is the only lowest grade: what is his maximum score?

For each player, the score S is the sum δ¹s¹+…+δ⁸s⁸, where the first term is an indicator for a correct answer and the second term is the sum over all other players of their complementary indicator, which can be replaced with the sum over all players since δ¹(1-δ¹)=0. Leading to the vector of scores

worz <- function(ansz){
  scor=apply(1-ansz,2,sum)
  return(apply(t(ansz)*scor,2,sum))}

Now, running by brute-force a massive number of simulations confirmed my intuition that the minimal winning score is 39, the number of players minus one [achieved by Alice giving a single good answer and the others none at all], while the maximum loosing score appeared to be 34, for which I had much less of an intuition!  I would have rather guessed something in the vicinity of 80 (being half of the answers replied correctly by half of the players)… Indeed, while in SIngapore, I however ran in the wee hours a quick simulated annealing code from this solution and moved to 77.

And the 2018 version of Le Monde maths puzzle competition starts today!, for a total of eight double questions, starting with an optimisation problem where the adjacent X table is filled with zeros and ones, trying to optimise (max and min) the number of positive entries [out of 45] for which an even number of neighbours is equal to one. On the represented configuration, green stands for one (16 ones) and P for the positive entries (31 of them). This should be amenable to a R resolution [R solution], by, once again!, simulated annealing. Deadline for the reply on the competition website is next Tuesday, midnight [UTC+1]

Le Monde puzzle [#1063] and le bac’

As pointed out by Jean-Louis Foulley in his comment on the recent Le Monde puzzle [#1063] post, the questions are heavily inspired by the 2018 high school final exam (known as Le Baccalauréat or Le Bac’) in mathematics in France, as can be checked from the above. Including the denomination of powerful number! In Part A, the students had to study the Diophantine equation x²-8y²=1 (E) and show it had an infinite number of (integer) solutions. While these students did not have access to computing facilities, question 1 was solved by sheer enumeration, providing 8=2³ and 9=3² as an answer. While (2) and (3) are rather straightforward, using the prime number decomposition of a and b for (2), the fact that x² and 8y² are powerful for (3). Brute force search leading to x=99,y=35 for the final question.

Le Monde puzzle [#1063]

A simple (summertime?!) arithmetic Le Monde mathematical puzzle

1. A “powerful integer” is such that all its prime divisors are at least with multiplicity 2. Are there two powerful integers in a row, i.e. such that both n and n+1 are powerful?
   
2.  Are there odd integers n such that n² – 1 is a powerful integer ?

The first question can be solved by brute force.  Here is a R code that leads to the solution:

isperfz <- function(n){ 
  divz=primeFactors(n) 
  facz=unique(divz) 
  ordz=rep(0,length(facz)) 
  for (i in 1:length(facz)) 
    ordz[i]=sum(divz==facz[i]) 
  return(min(ordz)>1)}

lesperf=NULL
for (t in 4:1e5)
if (isperfz(t)) lesperf=c(lesperf,t)
twinz=lesperf[diff(lesperf)==1]

with solutions 8, 288, 675, 9800, 12167.

The second puzzle means rerunning the code only on integers n²-1…

[1] 8
[1] 288
[1] 675
[1] 9800
[1] 235224
[1] 332928
[1] 1825200
[1] 11309768

except that I cannot exceed n²=10⁸. (The Le Monde puzzles will now stop for a month, just like about everything in France!, and then a new challenge will take place. Stay tuned.)

Le Monde puzzle [#1062]

A simple Le Monde mathematical puzzle none too geometric:

1. Find square triangles which sides are all integers and which surface is its perimeter.
2. Extend to non-square rectangles.

No visible difficulty by virtue of Pythagore’s formula:

for (a in 1:1e4)
for (b in a:1e4)
  if (a*b==2*(a+b+round(sqrt(a*a+b*b)))) print(c(a,b))

produces two answers

 5 12
 6  8

and in the more general case, Heron’s formula to the rescue!,

for (a in 1:1e2)
for (b in a:1e2)
for (z in b:1e2){
  s=(a+b+z)/2
  if (abs(4*s-abs((s-a)*(s-b)*(s-z)))<1e-4) print(c(a,b,z))}

returns

 4 15 21
 5  9 16
 5 12 13
 6  7 15
 6  8 10
 6 25 29
 7 15 20
 9 10 17

Le Monde puzzle [#1061]

A griddy Le Monde mathematical puzzle:

1. On a 4×5 regular grid, find how many nodes need to be turned on to see all 3×4 squares to have at least one active corner in case of one arbitrary node failing.
   
2.  Repeat for a 7×9 grid.
   

The question is open to simulated annealing, as in the following R code:

n=3;m=4;np=n+1;mp=m+1

cvr=function(grue){
  grud=grue
  obj=(max(grue)==0)
  for (i in (1:length(grue))[grue==1]){
   grud[i]=0
   obj=max(obj,max((1-grud[-1,-1])*(1-grud[-np,-mp])*
       (1-grud[-np,-1])*(1-grud[-1,-mp])))
   grud[i]=1}
  obj=99*obj+sum(grue)
  return(obj)}

dumban=function(grid,T=1e3,temp=1,beta=.99){
   obj=bez=cvr(grid)
   sprk=grid
   for (t in 1:T){
     grue=grid
     if (max(grue)==1){ grue[sample(rep((1:length(grid))[grid==1],2),1)]=0
      }else{ grue[sample(1:(np*mp),np+mp)]=1}
     jbo=cvr(grue)
     if (bez>jbo){ bez=jbo;sprk=grue}
     if (log(runif(1))<(obj-jbo)/temp){
        grid=grue;obj=cvr(grid)}
     temp=temp*beta
     }
   return(list(top=bez,sol=sprk))}

leading to

>  dumban(grid,T=1e6,temp=100,beta=.9999)
$top
[1] 8
$sol
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    1    0    1    0
[2,]    0    1    0    1    0
[3,]    0    1    0    1    0
[4,]    0    1    0    1    0

which sounds like a potential winner.