operation precisely impossible

Since the solution to the previous riddle from The Riddler on the maximum of  different terms in the composed operation

a∅b∅c∅d∅e∅f

depending on the bracketing ordering and the meaning of each ∅ among one of the six elementary operations got posted today as 974,860, I got back to my R code to understand why it differed from my figures by two orders of magnitude and realised I was overly trusting the R function unique. As it was returning more “different” entries than it should have, especially when choosing the six starting numbers (a,…,f) as Uniform (0,1). Using integers instead led for instance to 946,558, which was not so far from the target. But still imprecise as to whether or not some entries had been counted several times. I mentioned the issue to Robin, who rose to the challenge and within minutes came up with using the R function almost.unique from the CRAN package bazar, then producing outcomes like 974,513, hence quite close to 974,860 for random initialisations!

operation impossible

A riddle from The Riddler on how many different numbers one could at most produce from six initial values and the four basic operations. In other words, how many values could the terms in

a∅(b∅{c∅[d∅(e∅f)]})

and other positioning of the brackets could take? (With each ∅ being one of the four operations and a,…,f the initial values or a permutation of these.) A very crude evaluation leads to twenty million possible values, forgetting that addition and multiplication are commutative, while subtraction and division are anti-commutative. I tried a brute force approach in R, rather than exploring the tree of possible paths, but could not approach this figure by far, the number of different values still increasing for the largest manageable number of replicas I could try. Reducing the initial values at n=3, I could come closer to 123 with 95 different values and, for n=4, not too far from 1972 with 1687 values. Moving to n=6 by hopefully exhausting all (?) entries led to another guess of 50,524,809 values. I am however unsure that the R code is able to spot identical values, given the variability of the above figure…

Le Monde puzzle [#1105]

Another token game as Le Monde mathematical puzzle:

Archibald and Beatrix play with a pile of n>100 tokens, sequentially picking m tokens from the pile with m being a prime number [including m=1] or a multiple of 6, the winner taking the last tokens. If Beatrix knows n and proposes to Archibald to start, what is the value of n?

Which cannot be solved in a few lines of R code:

k<-function(n)n<4||all(n%%2:ceiling(sqrt(n))!=0)||!n%%6
g=(1:3)
n=c(4,i<-4)
while(max(n)<101){
  if(k(i)) g=c(g,i) else{
  while(i%in%g)i=i+1;j=4;o=!j
  while(!o&(j<i)){ 
    o=(j%in%n)&k(i-j);j=j+1}
  if(o) g=c(g,i) else n=c(n,i)}
  i=i+1}

since it returned no unsuccessful value above 100! With 4, 8, 85, 95, and 99 as predecessors. A rather surprising outcome and a big gap that most certainly has a straightforward explanation! Or a lack of understanding from yours truly: this post appears after the solution was published in Le Monde and I am more bemused than ever since the losing numbers in the journal are given as 4, 8, 85, … 89, and 129. With the slight hiccup that 89 is a prime number…. The other argument in the solution that there can only be five such losers is well-taken since there are only five possible non-zero remainders in the division by 6.

Le Monde puzzle [#1104]

A palindromic Le Monde mathematical puzzle:

In a monetary system where all palindromic amounts between 1 and 10⁸ have a coin, find the numbers less than 10³ that cannot be paid with less than three coins. Find if 20,191,104 can be paid with two coins. Similarly, find if 11,042,019 can be paid with two or three coins.

Which can be solved in a few lines of R code:

coin=sort(c(1:9,(1:9)*11,outer(1:9*101,(0:9)*10,"+")))
amounz=sort(unique(c(coin,as.vector(outer(coin,coin,"+")))))
amounz=amounz[amounz<1e3]

and produces 9 amounts that cannot be paid with one or two coins.

21 32 43 54 65 76 87 98 201

It is also easy to check that three coins are enough to cover all amounts below 10³. For the second question, starting with n¹=20,188,102,  a simple downward search of palindromic pairs (n¹,n²) such that n¹+n²=20,188,102 led to n¹=16,755,761 and n²=3,435,343. And starting with 11,033,011, the same search does not produce any solution, while there are three coins such that n¹+n²+n³=11,042,019, for instance n¹=11,022,011, n²=20,002, and n³=6.

Le Monde puzzle [#1099]

A simple 2×2 Le Monde mathematical puzzle:

Arielle and Brandwein play a game out of two distinct even integers between 1500 and 2500,  and y. Providing one another with either the pair (x/2,y+x/2) or the pair (x+y/2,y/2) until they run out of even possibilities or exceed 6 rounds. When x=2304, what is the value of y that makes Brandwein win?

Which I solved by a recursive function (under the constraint of a maximum of 11 levels of recursion):

nezt=function(x,y,i=1){
  if ((i>11)||((is.odd(x)&is.odd(y)))){ return(-1)
  }else{
    z=-1
    if (is.even(x)) z=-nezt(x/2,y+x/2,i+1)
    if (is.even(y)) z=max(z,-nezt(y/2,x+y/2,i+1))
    return(z)}}

and checking all values of y between 1500 and 2500 when x=2304, which produces y=1792 as the only value when Arielle loses. The reason behind (?) is that both 2304 and 1792 are divisible by 2⁸, which means no strategy avoids reaching stalemate after 8 steps, when it is Arielle’s turn to play.