(This article was first published on

**YGC » R**, and kindly contributed to R-bloggers)The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property. Let d_{1}be the 1st digit, d_{2}be the 2nd digit, and so on. In this way, we note the following: d_{2}d_{3}d_{4}=406 is divisible by 2 d_{3}d_{4}d_{5}=063 is divisible by 3 d_{4}d_{5}d_{6}=635 is divisible by 5 d_{5}d_{6}d_{7}=357 is divisible by 7 d_{6}d_{7}d_{8}=572 is divisible by 11 d_{7}d_{8}d_{9}=728 is divisible by 13 d_{8}d_{9}d_{10}=289 is divisible by 17 Find the sum of all 0 to 9 pandigital numbers with this property.

^{?}View Code RSPLUS

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rmDup <- function(vec) { idx <- sapply(vec, function(n) { nv <- unlist(strsplit(as.character(n), split="")) return(length(unique(nv)) == length(nv)) } ) return(vec[idx]) } n <- 102:999 prime <- c(13,11,7,5,3,2) d <- n[ n %% 17 ==0] d <- rmDup(d) retain <- c() for (i in 1:length(prime)) { for (j in d) { lastdigits <- j %% 10^i first2digit <- (j-lastdigits)/10^i for (n in 0:9) { m <- n*100+first2digit if( m %% prime[i] ==0 ) { retain <- c(retain,m*10^i+lastdigits) } } } d <- rmDup(retain) if (i != length(prime)) retain <- c() } s <- 0 for (i in d) { if(nchar(as.character(i)) == 9) { xx <- 0:9 firstDigit <- xx[!xx %in% unlist(strsplit(as.character(i), split=""))] s <- s+ firstDigit*10^9+i } if(nchar(as.character(i)) == 8) { xx <- 1:9 firstDigit <- xx[!xx %in% unlist(strsplit(as.character(i), split=""))] if(length(firstDigit) == 1) s <- s+ firstDigit*10^9+i } } print(s) |

The implementation is not elegant, but amazingly fast.

> system.time(source("Problem43.R")) [1] 16695334890 user system elapsed 0 0 0

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