## Le Monde puzzle [#1127]

Posted in Books, Kids, R, Statistics with tags , , , , on January 17, 2020 by xi'an

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!

## Le Monde puzzle [#1120]

Posted in Books, Kids, pictures, R with tags , , , , on January 14, 2020 by xi'an

A board game as Le weekly Monde current mathematical puzzle:

11 players in a circle and 365 tokens first owned by a single player. Players with at least two tokens can either remove one token and give another one left or move two right and one left. How quickly does the game stall, how many tokens are left, and where are they?

The run of a R simulation like

od=function(i)(i-1)%%11+1
muv<-function(bob){
if (max(bob)>1){
i=sample(rep((1:11)[bob>1],2),1)
dud=c(0,-2,1)
if((runif(1)<.5)&(bob[i]>2))dud=c(2,-3,1)
bob[c(od(i+10),i,od(i+1))]=bob[c(od(i+10),i,od(i+1))]+dud
}
bob}

always provides a solution

> bob
[1] 1 0 1 1 0 1 1 0 1 0 0


with six ones at these locations. However the time it takes to reach this frozen configuration varies, depending on the sequence of random choices.

## 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
[1] 1 2
[1] 1 4
[1] 2 3
[1] 3 4


## Le Monde puzzle [#1121]

Posted in Books, Kids with tags , , , , on December 17, 2019 by xi'an

A combinatoric puzzle as Le weekly Monde current mathematical puzzle:

A class of 75<n<100 students is divided at random into two groups of sizes a and b=n-a, respectively, such that the probability that two particular students Ji-ae and Jung-ah have a probability of exactly 1/2 to be in the same group. Find a and n.

(with an original wording mentioning an independent allocation to the group!). Since the probability to be in the same group (under a simple uniform partition distribution) is

$\frac{a(1-1)}{n(n-1)}+\frac{b(b-1)}{n(n-1)}$

it is sufficient to seek by exhaustion values of (a,b) such that this ratio is equal to ½. The only solution within the right range is then (36,45) (up to the symmetric pair). This can be also found by seeking integer solutions to the second degree polynomial equation, namely

$b^\star=\left[ 1+2a\pm\sqrt{1+8a}\right]/2 \in \mathbb N$

## Le Monde puzzle [#1119]

Posted in Kids, R with tags , , , , on December 8, 2019 by xi'an

A digit puzzle as Le weekly Monde current mathematical puzzle that sounds close to some earlier versions:

Perfect squares are pairs (a²,b²) with the same number of digits such that a²b² is itself a square. What is the pair providing a²b² less than 10⁶? Is there a solution with both integers enjoying ten digits?

The run of a brute force R code like

cek<-function(a,b){
u<-trunc
if ((n<-u(log(a^2,ba=10)))==u(log(b^2,ba=10))&
(u(sqrt(a^2*10^(n+1)+b^2))^2==(a^2*10^(n+1)+b^2))) print(c(a,b))}

provides solutions to the first question.

[1] 2 3
[1] 4 9
[1] 12 20
[1] 15 25
[1] 18 30
[1] 49 99
[1] 126 155
[1] 154 300
[1] 159 281
[1] 177 277
[1] 228 100
[1] 252 310
[1] 285 125


with the (demonstrable) conclusion that the only pairs with an even number of digits are of the form (49…9²,9…9²), as for instance (49999²,99999²) with ten digits each.

## Froebenius coin problem

Posted in pictures, R, Statistics with tags , , , , , , , , , , on November 29, 2019 by xi'an

A challenge from The Riddler last weekend came out as the classical Frobenius coin problem, namely to find the largest amount that cannot be obtained using only n coins of specified coprime denominations (i.e., with gcd equal to one). There is always such a largest value. For the units a=19 and b=538, I ran a basic R code that returned 9665 as the largest impossible value, which happens to be 19×538-538-19, the Sylvester solution to the problem when n=2. A recent paper by Tripathi (2017) manages the case n=3, for “almost all triples”, which decomposes into a myriad of sub-cases. (As an aside, Tripathi (2017) thanks a PhD student, Prof. Thomas W. Cusick, for contributing to the proof, which constitutes a part of his dissertation, but does not explain why he did not join as co-author.) The specific case when a=19, b=101, and c=538 suggested by The Riddler happens to fall in one of the simplest categories since, as ⌊cb⁻¹⌋ and ⌊cb⁻¹⌋ (a) are equal and gcd(a,b)=1 (Lemma 2), the solution is then the same as for the pair (a,b), namely 1799. As this was quite a light puzzle, I went looking for a codegolf challenge that addressed this problem and lo and behold! found one. And proposed the condensed R function

function(a)max((1:(b<-prod(a)))[-apply(combn(outer(a,0:b,"*"),sum(!!a))),2,sum)])

that assumes no duplicate and ordering in the input a. (And learned about combn from Robin.) It is of course very inefficient—to the point of crashing R—to look at the upper bound

$\prod_{i=1}^n a_i \ \ \ \ \ \ \ (1)$

for the Frobenius number since

$\min_{(i,j);\text{gcd}(a_i,a_j)=1} (a_i-1)(a_j-1)\ \ \ \ \ \ \ (2)$

is already an upper bound, by Sylvester’s formula. But coding (2) would alas take much more space…

## riddle of the seats

Posted in Statistics with tags , , , on November 8, 2019 by xi'an

An arithmetic quick riddle from The Riddler:

If an integer n is a multiple of every integer between 1 and 200, except for two consecutive ones, find those consecutive integers.

Since the highest power of 2 less than 200 is 2⁷=128 and since 127 is a prime number, the number

$2^6\times \prod_{i=0,i\ne 63}^{99} (2i+1)$

should work in that it contains all odd integers but 127, and all even numbers, but 128. Of course a smaller number that avoids duplicates by only considering the 44 primes other than 127 and 2 to a power that keep them less than 200 is also valid. Which gives a number of the order of 1.037443 10⁸⁵.