Archive for brute force

Le Monde puzzle [#1105]

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

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:

  if(k(i)) g=c(g,i) else{
  if(o) g=c(g,i) else n=c(n,i)}

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]

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

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:


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]

Posted in Books, Kids, R with tags , , , , , on April 28, 2019 by xi'an

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):

  if ((i>11)||((is.odd(x)&is.odd(y)))){ return(-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))

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.

Le Monde puzzle [#1094]

Posted in Books, Kids, R with tags , , , , , , on April 23, 2019 by xi'an

A rather blah number Le Monde mathematical puzzle:

Find all integer multiples of 11111 with exactly one occurrence of each decimal digit..

Which I solved by brute force, by looking at the possible range of multiples (and  borrowing stringr:str_count from Robin!)

> combien=0
> for (i in 90001:900008){
> combien
[1] 3456

And a bonus one:

Find all integers y that can write both as x³ and (10z)³+a with 1≤a≤999.

which does not offer much in terms of solutions since x³-v³=(x-v)(x²+xv+v²)=a shows that x² is less than 2a/3, meaning x is at most 25. Among such numbers only x=11,12 lead to a solution as x³=1331,1728.

Le Monde puzzle [#1092]

Posted in Statistics with tags , , , , , , , on April 18, 2019 by xi'an

A Latin square Le Monde mathematical puzzle that I found rather dreary:

A hidden 3×3 board contains all numbers from 1 to 9. Anselm wants to guess the board and makes two proposals. Berenicke tells him how many entries are in the right rows and colums for each proposal, along with the information that no entry is at the right location. Anselm deduces the right board.

Which I solved by brute force and not even simulated annealing, first defining a target


on a 9×9 board entry reproducing all items of information given by Berenicke. If all constraints are met, the function returns 14. And then searched for a solution at random:

  for (t in 1:1e6){
    if (fitz(ordo)==14){

which produces the correct board

4 3 5
6 7 1
9 2 8