**A**bout this #904 arithmetics Le Monde mathematical puzzle:

Find all pluralintegers, namely positiveintegers such that (a) none of their digits is zero and (b) removing their leftmost digit produces a dividing plural integer (with the convention that one digit integers are all plural).

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) [1] 11 12 15 21 22 24 25 31 32 33 35 36 [13] 41 42 44 45 48 51 52 55 61 62 63 64 [25] 65 66 71 72 75 77 81 82 84 85 88 91 [37] 92 93 95 96 99 125 225 312 315 325 375 425 [49] 525 612 615 624 625 675 725 735 825 832 912 [61] 915 925 936 945 975 1125 2125 3125 3375 4125 [70] 5125 5625 [72] 6125 6375 7125 8125 9125 9225 9375 53125 [80] 91125 95625