# project euler-Problem 43

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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 d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following: d2d3d4=406 is divisible by 2 d3d4d5=063 is divisible by 3 d4d5d6=635 is divisible by 5 d5d6d7=357 is divisible by 7 d6d7d8=572 is divisible by 11 d7d8d9=728 is divisible by 13 d8d9d10=289 is divisible by 17 Find the sum of all 0 to 9 pandigital numbers with this property.

^{?}View Code RSPLUS

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 |
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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