Le Monde puzzle [#913]

An arithmetics Le Monde mathematical puzzle:

Find all bi-twin integers, namely positive integers such that adding 2 to any of their dividers returns a prime number.

An easy puzzle, once the R libraries on prime number decomposition can be found!, since it is straightforward to check for solutions. Unfortunately, I could not install the recent numbers package. So I used instead the schoolmath R package. Despite its possible bugs. But it seems to do the job for this problem:

lem=NULL
for (t in 1:1e4) 
  if (prod(is.prim(prime.factor(t)+2)==1)) 
    lem=c(lem,t)digin=function(n){

which returned all solutions, albeit in a lengthy fashion:

> lem
 [1] 1 3 5 9 11 15 17 25 27 29 33 41 45 51 55
 [16] 59 71 75 81 85 87 99 101 107 121 123 125 135 137 145
 [31] 149 153 165 177 179 187 191 197 205 213 225 227 239 243 255
 [46] 261 269 275 281 289 295 297 303 311 319 321 347 355 363 369
 [61] 375 405 411 419 425 431 435 447 451 459 461 493 495 505 521
 [76] 531 535 537 561 569 573 591 599 605 615 617 625 639 641 649
 [91] 659 675 681 685 697 717 725 729 745 765 781 783 807 809 821
[106] 825 827 841 843 857 867 881 885 891 895 909 933 935 955 957
[121] 963 985 1003 1019 1025 1031 ...

Le Monde puzzle [#857]

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 [#6]

A simple challenge in Le Monde this week: find the group of four primes such that any sum of three terms in the group is prime and the overall sum is minimised. Here is a quick exploration by simulation, using the schoolmath package (with its imperfections):

A=primes(start=1,end=53)[-1]
lengthA=length(A)

res=4*53
for (t in 1:10^4){

 B=sample(A,4,prob=1/(1:lengthA))
 sto=is.prim(sum(B[-1]))
 for (j in 2:4)
 sto=sto*is.prim(sum(B[-j]))

 if ((sto)&(sum(B)<res)){
 res=sum(B)
 sol=B}
 }
}

providing the solution 5 7 17 19.

A subsidiary question in the same puzzle is whether or not it is possible to find a group of five primes such that any sum of three terms is still prime. Running the above program with the proper substitutions of 4 by 5 does not produce any solution, even when increasing the upper boundary in A. So it is most likely that the answer is no.

Le Monde puzzle [41]

The current puzzle in Le Monde this week is again about prime numbers:

The control key on a credit card is an integer η(a) associated with the card number a such that, if the card number is c=ab, its key η(c) satisfies η(c)=η(a)+η(b)-1. There is only one number with a key equal to 1 and the keys of 160 and 1809 are 10 and 7, respectively. What is the key of 2010?

The key of 1 is necessarily 1 since

η(a1)=η(a)+η(1)-1=η(a).

So this eliminates 1. Now, the prime number decompositions of 160, 1809, and 2010 are given by

> prime.factor(160)
[1] 2 2 2 2 2 5
> prime.factor(1809)
[1]  3  3  3 67
> prime.factor(2010)
[1]  2  3  5 67

using a function of the (still bugged!) schoolmath package. We thus have the decompositions

η(160)+5=5η(2)+η(5)=15

η(1809)+3=3η(3)+η(67)=10

Since η(2) cannot be 1 and is an integer, we necessarily have η(2)=2 and this implies η(5)=5. With the same constraint on 1, the second sum also leads to the unique solution η(3)=2 and η(67)=4. The solution is thus η(2010)=13-3=13. (For ‘Og readers who saw several versions of this post, the subsidiary question is how many versions was there?!)

bug in schoolmath

Neil Gunther has pointed out on his blog that the prime number decomposition R package schoolmath contains mistakes in the function primes, listing 1 as a prime number but also including decomposable numbers like 133 in its list of prime numbers:

> primes(100,140)
[1] 101 107 111 113 123 129 131 137
> primes(50,140)
[1]  51  53  59  61  67  71  73  79  83  89  97 101 103 107 109 113 127 131 133
[20] 137 139
> is.prim(primes(133)
[1] FALSE
> is.prim(primes(200,300))
[1] FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE
[13] TRUE TRUE TRUE TRUE TRUE TRUE
> sum(1-is.prim(primes(1,1000)))
[1] 10
> data(primlist)
> sum(1-is.prim(primlist[1:25000]))
[1] 3309

This is rather annoying and I hope it gets quickly fixed!