# project euler-Problem 43

November 7, 2011
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(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 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