Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. About this #904 arithmetics Le Monde mathematical puzzle:

Find all plural integers, namely positive integers such that (a) none of their digits is zero and (b) removing their leftmost digit produces a dividing integer.

a slight modification in the R code allows for a faster exploration, based on the fact that solutions add one extra digit to solutions with one less digit:

First, I found this function on Stack Overflow to turn an integer into its digits:

```pluri=plura=NULL
#solutions with two digits
for (i in 11:99){

dive=rev(digin(i)[-1])
if (min(dive)>0){
dive=sum(dive*10^(0:(length(dive)-1)))
if (i==((i%/%dive)*dive))
pluri=c(pluri,i)}}

for (n in 2:6){ #number of digits
plura=c(plura,pluri)
pluro=NULL
for (j in pluri){

for (k in (1:9)*10^n){
x=k+j
if (x==(x%/%j)*j)
pluro=c(pluro,x)}
}
pluri=pluro}
```

which leads to the same output

```> sort(plura)
 11 12 15 21 22 24 25 31 32 33 35 36
 41 42 44 45 48 51 52 55 61 62 63 64
 65 66 71 72 75 77 81 82 84 85 88 91
 92 93 95 96 99 125 225 312 315 325 375 425
 525 612 615 624 625 675 725 735 825 832 912
 915 925 936 945 975 1125 2125 3125 3375 4125
 5125 5625
 6125 6375 7125 8125 9125 9225 9375 53125
 91125 95625
```

Filed under: Books, Kids, R, Statistics, University life Tagged: arithmetics, Le Monde, mathematical puzzle, strsplit()  