Le Monde puzzle [38]

Since I have resumed my R class, I will restart my resolution of Le Monde mathematical puzzles…as they make good exercises for the class. The puzzle this week is not that exciting:

Find the four non-zero different digits a,b,c,d such that abcd is equal to the sum of all two digit numbers made by picking without replacement two digits from {a,b,c,d}.

The (my) dumb solution is to proceed by enumeration

for (a in 1:9){
for (b in (1:9)[-a]){
for (c in (1:9)[-c(a,b)]){
for (d in (1:9)[-c(a,b,c)]){
 if (231*sum(c(a,b,c,d))==sum(10^(0:3)*c(a,b,c,d)))
    print(c(a,b,c,d))
}}}}

taking advantage of the fact that the sum of all two-digit numbers is (30+4-1) times the sum a+b+c+d, but there is certainly a cleverer way to solve the puzzle (even though past experience has shown that this was not always the case!)

3 Responses to “Le Monde puzzle [38]”

  1. [...] Monde puzzle [34] Since the puzzle in this week (-end) edition of Le Monde is not (easily) solvable via an R program, I chose to go [...]

  2. Not much different but checks only a small set..

    library(gregmisc)
    mysets <-permutations(9, 4)
    universe.sol<-mysets%*%10^(3:0)
    possible.sol<-intersect(7*11*3*c(10:30),universe.sol)

    # now check these 7 possibilities
    for(i in 1:length(possible.sol)){
    if(possible.sol[i]==231*sum(as.numeric(unlist(strsplit(as.character(possible.sol[i]),NULL)))))
    print(possible.sol[i])
    }

  3. [...] on PIN number By robinryder Now that the new school year has started, Christian Robert has picked up solving the Le Monde mathematical puzzles using R again, which leads me to solving them without [...]

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