**S**ince 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!)

Filed under: R, University life Tagged: class, Le Monde, mathematical puzzle, R

*Related*

To

**leave a comment** for the author, please follow the link and comment on his blog:

** Xi'an's Og » R**.

R-bloggers.com offers

**daily e-mail updates** about

R news and

tutorials on topics such as: visualization (

ggplot2,

Boxplots,

maps,

animation), programming (

RStudio,

Sweave,

LaTeX,

SQL,

Eclipse,

git,

hadoop,

Web Scraping) statistics (

regression,

PCA,

time series,

trading) and more...

If you got this far, why not

__subscribe for updates__ from the site? Choose your flavor:

e-mail,

twitter,

RSS, or

facebook...

**Tags:** class, Le Monde, mathematical puzzle, R, University life