Archive for number theory

Le Monde puzzle [#1013]

Posted in Books, Kids with tags , , , , , on June 23, 2017 by xi'an

A purely arithmetic Le Monde mathematical puzzle:

An operation þ applies to all pairs of natural integers with the properties

0 þ (a+1) = (0 þ a)+1, (a+1) þ (b+1)=(a þ b)+1, 271 þ 287 = 77777, 2018 þ 39 = 2018×39

Find the smallest integer d>287 such that there exists c<d leading to c þ d = c x d, the smallest integer f>2017 such that 2017 þ f = 2017×40. Is there any know integer f such that f þ 2017 = 40×2017?

The major appeal in this puzzle (where no R programming seems to help!) is that the “data” does not completely defines the operation  þ ! Indeed, when a<b, it is straightforward to deduce that a þ b = (0 þ 0)+b, hence solving the first two questions by deriving (0 þ 0)=270×287 [with d=315 and f=2017×40-270×287], but the opposed quantity b þ a is not defined, apart from (2018-39) þ 0. This however brings a resolution since

(2018-39) þ 0 = 2017×39 and (2018-39+2017) þ 2017 = 2017×39+2017 = 2017×40

leading to f=2018-39+2017=3996.

Le Monde puzzle [#960]

Posted in Kids, R with tags , , , , , on April 28, 2016 by xi'an

An arithmetic Le Monde mathematical puzzle:

Given an integer k>1, consider the sequence defined by F(1)=1+1 mod k, F²(1)=F(1)+2 mod k, F³(1)=F²(1)+3 mod k, &tc. [With this notation, F is not necessarily a function.] For which value of k is the sequence the entire {0,1,…,k-1} set?

This leads to an easy brute force resolution, for instance writing the R function

crkl<-function(k) return(unique(cumsum(1:(2*k))%%k))

where 2k is a sufficient substitute for ∞. Then the cases where the successive images of 1 visit the entire set {0,1,…,k-1} are given by

> for (i in 2:550) if (length(crkl(i))==i) print(i)
[1] 2
[1] 4
[1] 8
[1] 16
[1] 32
[1] 64
[1] 128
[1] 256
[1] 512

which suspiciously looks like the result that only the powers of 2 k=2,2²,2³,… lead to a complete exploration of the set {0,1,…,k-1}. Checking a few series in the plane back from Warwick, I quickly found that when k is odd, (1) the sequence is of period k and (2) there is symmetry in the sequence, which means it only takes (k-1)/2 values. For k even, there is a more complicated symmetry, with the sequence being of period 2k, symmetric around its two middle values, and taking the values 1,2,..,1+k(2k+1)/4,..,1+k(k+1)/2. Those values cannot cover the set {0,1,…,k-1} if two are equal, which means an i(i+1)/2 congruent to zero modulo k, hence equal to k. This is clearly impossible when k is a power of 2 because i and i+1 cannot both divide a power of 2. I waited for the published solution as of yesterday’s and the complete argument was to show that when N=2p, the corresponding sequence [for N] is made (modulo p) of the sequence for p plus the same sequence translated by p. The one for N is complete only if the one for p is complete, which by recursion eliminates all cases but the powers of 2…

Le Monde puzzle [#857]

Posted in Books, Kids, R with tags , , , , , , , on March 22, 2014 by xi'an

A rather bland case of Le Monde mathematical puzzle :

Two positive integers x and y are turned into s=x+y and p=xy. If Sarah and Primrose are given S and P, respectively, how can the following dialogue happen?

  • I am sure you cannot find my number
  • Now you told me that, I can, it is 46.

and what are the values of x and y?

In the original version, it was unclear whether or not each person knew she had the sum or the product. Anyway, the first person in the dialogue has to be Sarah, since a product p equal to a prime integer would lead Primrose to figure out x=1 and hence s=p+1. (Conversely, having observed the sum s cannot lead to deduce x and y.) This means x+y-1 is not a prime integer. Now the deduction of Primrose that the sum is 46 implies p can be decomposed only once in a product such that x+y-1 is not a prime integer. If p=45, this is the case since 45=15×3 and 45=5×9 lead to 15+3-1=17 and 5+9-1=13, while 45=45×1 leads to 45+1-1=45.  Other solutions fail, as demonstrated by the R code:

 > for (x in 1:23){
 + fact=c(1,prime.factor(x*(46-x)))
 + u=0;
 + for (i in 1:(length(fact)-1))
 + u=u+1-is.prim(prod(fact[1:i])+prod(fact[-(1:i)])-1)
 + if (u==1) print(x)}
 [1] 1

Busser and Cohen argue much more wisely in their solution that any non-prime product p other than 45 would lead to p+1 as an acceptable sum s, hence would prevent Primrose from guessing s.

Le Monde puzzle [#839]

Posted in Books, Kids, R with tags , , on November 16, 2013 by xi'an

A number theory Le Monde mathematical puzzle whose R coding is not really worth it (and which rings a bell of a similar puzzle in the past, puzzle I cannot trace…):

The set Ξ is made of pairs of integers (x,y) such that (i) both x and y are written as a sum of two squared integers (i.e., are bisquare numbers) and (ii) both xy and (x+y) are bisquare numbers. Why is the product condition superfluous?  For which values of (a,b) is the pair (13a,13b) in Ξ ?

In the first question, the property follows from the fact that the product of two bisquare numbers is again a bisquare number, thank to the remarkable identity

(a²+b²)(c²+d²) = (ac+bd)²+(ad-bc)²

(since the double products cancel). For the second question, once I realised that


it followed that any number 13a was the sum of two squares, hence a bisquare number, and thus that the only remaining constraint was that (b≥a)


is also bisquare. If b-a is even, this sum is then the product of two bisquare numbers and hence a bisquare number. If b-a is odd, I do not have a general argument to bar the case (it certainly does not work for 13+13² and the four next ones).

Le Monde puzzle [#818]

Posted in Books, Kids, R with tags , , , , on May 1, 2013 by xi'an

The current puzzle is as follows:

Define the symmetric of an integer as the integer obtained by inverting the order of its digits, eg 4321 is the symmetric of 1234. What are the numbers for which the square is equal to the symmetric of the square of the symmetric?

I first consulted stackexchange to find a convenient R function to create the symmetric:

  function(y) substr(x, y, y)))}



and then found that all integers but the multiples of 10 are symmetric! only some integers involving 1,2,3 and sometimes zero would work:

> for (i in 1:1000){
+   if (i^2==flip(flip(i)^2)) print(i)}
[1] 1
[1] 2
[1] 3
[1] 11
[1] 12
[1] 13
[1] 21
[1] 22
[1] 31
[1] 101
[1] 102
[1] 103
[1] 111
[1] 112
[1] 113
[1] 121
[1] 122
[1] 201
[1] 202
[1] 211
[1] 212
[1] 221
[1] 301
[1] 311

If I am not (again) confused, the symmetric integers would be those (a) not ending with zero and (b) only involving digits whose products are all less than 10.

tak1ng sudoku ser1ously

Posted in Books, Statistics, University life with tags , , , , , on March 14, 2013 by xi'an

“There is something deeply satisfying in encountering opacity.” (p.9)

I think it was last summer at the Australasian Statistics conference in Adelaide that I saw this book by Jason Rosenhouse and Laura Taalman, Taking Sudoku seriously: The math behind the World’s most popular pencil puzzle. (Or was it in Kyoto at the ISBA meeting?!) In any case, I mentioned my long-term interest in this puzzle (as readers of the ‘Og will undoubtedly have noticed!) and proposed to write a review for Chance. A few weeks later the book arrived at home. Which was a mistake as my wife who is a much deeper Sudoku fanatic than I stole the book with a vague promise to write the review! It is only a few weeks ago that I was able to truly get the book (if not the review) back…

“This is surely the most trivial of pursuits.” (p.4)

In case you never tried to solve a Sudoku puzzle, open your favourite newspaper, news website or phone, and you should immediately find one of those 9×9 squares to fill with integers between 1 and 9, under three sets of constraints: row-wise, column-wise, and block-wise. (The book describes all kinds of variations with no particularly added value.) Depending on the number of integers already present on the grid and on their location, the resolution of the puzzle gets from obvious to close to impossible by hand. This range in the difficulty and the resulting challenges may explain for the popularity of the method, although it remains a mystery to many bystanders. There are many strategies for solving Sudokus, from the (Sherlock Holmes) elimination of obviously impossible values to branching processes carrying guesses until a binary conclusion. The book covers some of those: forced cells, twins, X-wings, and Ariadne’s thread (moving from Star Wars to The Lord of the Ring!). It however does not bring any novel technique to solve Sudokus, at least at the pencil level, so potential readers with this intent in mind should abstain!

“Sudoku is math in the small.” (p.171)

Now many may wonder about the connections between Sudokus and mathematics. Some mathematicians are clearly sceptical. For instance, I remember discussing solving Sudoku grids in the métro with a friend and Mathematics colleague from Paris-Dauphine and she was quite surprised by this fascination, as she did not see any mathematical appeal in the exercise. In my eyes, this is a logical puzzle at the very least (there is one way of setting all the entries and some obligatory paths for filling the grid that have to be exposed), which makes Sudoku vaguely mathematical. I am also attracted by the automated resolution of those puzzles, as shown by my simulated annealing attempt a (long) while ago, even though I am aware that the state-of-the-art codes are beyond my reach. (They are not detailed either in the book.) Reading this book actually made me think of potential alternatives for simulating annealing, from using additional integers during relaxation, to solving a set of algebraic equations under relaxed constraints. (I wonder if Gröbner bases are used by the fastest resolution codes.)

“We are unaware of any result about Sudoku puzzles provable using the techniques of graph-coloring that could not have been arrived at more easily by other means.” (p.133)

Taking Sudoku Seriously does exhibit some of those connections, it may however feel contrived for some readers (and unintelligible to others), as the mathematical objects have an existence of their own and do not appear to directly contribute to the resolution of the puzzle. An exception in the first case is the one of Latin and Greco-Latin squares in Chapters 2 and 4, which are Sudokus of sort, while an exception to the second is the mention of Gröbner bases in Chapter 8 for solving sets of algebraic equations providing the solution to Sudokus in the form of complex roots of unity. Some number theory gets introduced along the way, as well as algebra (permutation and symmetry groups, polynomials) and graph theory. Analysis is rather absent from this volume and so is probability (hence simulated annealing). Except for a rather artificial and unrelated entry about Newton’s cannonball and Archimede’s derivation of the area of a circle in Chapter 7.

“We have a number of different audiences in mind.” (p.xi)

A few days prior to finish reading the book, I received my copy of Significance with a review of Taking Sudoku Seriously by Nicola Tilt. Glancing quickly at it, I read she was wondering about the intended audience for the book, which also is an issue with me. Neophytes and mathematicians alike will learn little about maths from reading the book, since the difficult questions (like counting the number of Sudokus and of non-equivalent Sudokus, or creating Sudokus) are not answered. The most interesting input is, in my opinion, the discussion of computer-based mathematical proofs. From the four-colour theorem to counting the number of Sudokus, to counting the number of fundamentally different Sudoku squares [as 5,472,730,538]. Unfortunately, the book publication date (April 2011) means that it missed the “big news”, namely the resolution of what the authors call “The rock star problem”, which is the determination of “the minimum number of starting clues possible in a Sudoku puzzle with a unique solution” (p.164). On Jan. 1, 2012, Gary McGuire, Bastian Tugemann, and Gilles Civario from University College Dublin used a mix of mathematics and computer power to prove that the minimum is indeed 17, as posited by the authors. (A result to feature prominently in the revised edition, if any!)

“We are curious for a living.” (p.191)

To conclude, and to be fair with the authors, the book reads nicely (on high quality glossy paper), once you limit your expectations to see some links between maths and Sudokus, it offers a lot of Sudokus and related puzzles, with solutions at the back of the book, and, above all, it teaches to some degree mathematical reasoning by running complete arguments on 4×4 Shidokus and extrapolating the results to the 9×9 Sudokus (or not, as in the case of polynomials where the extrapolation does not work, Section 8.2). Worth perusing on your way to work, trying to complete the puzzle at hand before the next stop!

As a postultimate remark, a paper by Hiroshi Watanabe from the University of Tokyo was arXived a few days ago: it contains further links about the mathematical properties of Sudokus, incl. one with a nine-state Potts model, but mostly a simulated annealing algorithm to find a “difficult” Sudoku. Contrary to my early attempt, this algorithm explores the space of all Sudoku puzzles, thus moves from one Sudoku to another at each iteration (“with the Metropolis criterion with the Boltzmann weight”, I wonder why Boltzmann?!).The hardest sudoku found by this method is represented below: it has a depth of 10 and would require 50,000 backtracking attempts to solve! Just “impossible to solve by hand”. (It took the equivalent of 11 cpu-core-years to create it.) As pointed out to me by Anthony Leverrier, from INRIA, the author added in the second version of his arXiv preprint that the puzzle can be easily solved by hand… So you may give it a try.

The hardest Sudoku ever, by H. Watanabe

the ABC conjecture

Posted in Books, University life with tags , , , , on November 27, 2012 by xi'an

Both Pour la Science and La Recherche, two French science magazines, had an entry this month on the abc conjecture! However, ABC being a common accronym, it is alas unrelated with my research theme. The abc conjecture is a number theory conjecture that states that if a and b are integers with no common factor and a small number of prime dividers, this does not hold for c=a+b. This is the abc triplet. (More precisely, the conjecture states that the quality of the triplet abc:

q(a,b,c) = \log c / \log \text{rad}(abc)

is larger than 1+ε for a finite number of triplets abc.) A proof of the conjecture by Shinichi Mochizuki was recently proposed, hence the excitment in the community. In La Recherche, I read that this conjecture is associated with an interesting computing challenge, namely to find the exhaustive collection of triplets with a quality more than a given bound 1+ε.