Le Monde puzzle [#954]

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A square Le Monde mathematical puzzle:

Given a triplet (a,b,c) of integers, with a? Can you find the triplet (a,b,c) that produces the sum a+b+c closest to 1000?

This is a rather interesting challenge and a brute force resolution does not produce interesting results. For instance, using the function is.whole from the package Rpmfr, the R functions

ess <- function(a,b,k){
#assumes a


 while (b<1000*a){
  if (is.whole(sqrt(a+b))){
   while (k<100*b){
    if (is.whole(sqrt(a+k))&is.whole(b+k))
     if (ess(a,b,k)) break()

do not return any solution when a=1,2,3,4,5

Looking at the property that a+b,a+c,b+c, and a+b+c are perfect squares α²,β²,γ², and δ². This implies that

a=(δ+β)(δ-β), b=(δ+γ)(δ-γ), and c=(δ+α)(δ-α)

with 1<α<β<γ<δ. If we assume β²,γ², and δ² consecutive squares, this means β=γ-1 and δ=γ+1, hence

a=4γ, b=2γ+1, and c=(γ+1+α)(γ+1-α)

which leads to only two terms to examine. Hence writing another R function


and running a check for the smallest values of α and γ leads to the few solutions available:

> for (ga in 3:1e4) 
for(al in 1:(ga-2)) 
if (ess(abc(al,ga))) print(abc(al,ga))
[1] 80 41 320
[1] 112 57 672
[1] 192 97 2112
[1] 240 121 3360
[1] 352 177 7392
[1] 416 209 10400
[1] 560 281 19040
[1] 640 321 24960
[1] 816 409 40800
[1] 912 457 51072

Filed under: Kids, R Tagged: Alice and Bob, is.whole, Le Monde, mathematical puzzle, R, recursive function, Rmpfr

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