Le Monde puzzle [#959]

Another of those arithmetic Le Monde mathematical puzzle:

Find an integer A such that A is the sum of the squares of its four smallest dividers (including1) and an integer B such that B is the sum of the third poser of its four smallest factors. Are there such integers for higher powers?

This begs for a brute force resolution checking the integers until a solution appears. The only exciting part is providing the four smallest factors but a search on Stack overflow led to an existing R function:

FUN <- function(x) {
    x <- as.integer(x)
    div <- seq_len(abs(x))
    return(div[x %% div == 0L])
}

(which uses the 0L representation I was unaware of) and hence my R code:

quest1<-function(n=2){
 I=4
 stop=TRUE
 while ((stop)&(I<1e6)){
  I=I+1
  dive=FUN(I)
  if (length(dive)>3)
     stop=(I!=sum(sort(dive)[1:4]^n))
   }
 return(I)
 }

But this code only seems to work for n=2 as produces A=130: it does not return any solution for the next value of n… As shown by the picture below, which solely exhibits a solution for n=2,5, A=17864 (in the second case), there is no solution less than 10⁶ for n=3,4,6,..9. So, unless I missed a point in the question, the solutions for n>2 are larger if they at all exist.

lemonde959A resolution got published yesterday night in Le Monde  and (i) there is indeed no solution for n=3 (!), (ii) there are solutions for n=4 (1,419,874) and n=5 (1,015,690), which are larger than the 10⁶ bound I used in the R code, (iii) there is supposedly no solution for n=5!, when the R code found that 17,864=1⁵+2⁵+4⁵+7⁵… It is far from the first time the solution is wrong or incomplete!

One Response to “Le Monde puzzle [#959]”

  1. […] Son unas entradas muy interesantes, porque siempre aprendo una función o un truco nuevos. En la entrada del pasado miércoles 20 de abril, Xian atacaba el problema 959, del fin de […]

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