# [Project Euler] – Problem 58

**YGC**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

**37** 36 35 34 33 32 **31**

38 **17** 16 15 14 **13** 30

39 18 **5 ** 4 **3** 12 29

40 19 6 1 2 11 28

41 20 **7** 8 9 10 27

42 21 22 23 24 25 26

**43** 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

^{?}View Code RSPLUS

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | side.length = 1 x <- 1 iter <- 1 ratio = 0.6 isp.n <- 0 N <- 0 while(ratio > 0.1) { last.x <- x[length(x)] side.length <- side.length + 2 x <- rep(last.x,4 ) + c(2,4,6,8)* iter iter <- iter + 1 isp <- gmp::isprime(x) isp.n <- isp.n + sum(as.logical(isp)) N <- N + 4 ratio <- isp.n/N print(side.length) } |

**Answer**: 26241

### Related Posts

**leave a comment**for the author, please follow the link and comment on their blog:

**YGC**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.