F# Returning a Boolean -

first shot @ euler #3 in f#, , return boolean more elegantly mutable value.

// number prime if can divide , 1.  can odd. let isprime x =     if (x%2l = 0l)         false     else         let mutable result = true         in 3l..x/2l             if (x%i = 0l)                 result <- false         result  let = isprime(17l)  // true printfn "%b" 

the l's i'm forcing function return bigints (there has better way too, 1 step @ time)....

edit gradbot's solution

let isprime x =     // prime number can't     if (x%2l = 0l)         false     else         // check divisors (other 1 , itself) half value of number eg 15 check 7         let maxi = x / 2l          let rec notdivisible =             // if we're reached more value check prime             if > maxi                 true             // found divisor false             elif x % = 0l                 false             // add 2 'loop' , call again             else                 notdivisible (i + 2l)          // start @ 3         notdivisible 3l 

you can replace else clause forall:

seq.forall (fun -> x % <> 0l) { 3l .. x/2l } 

and further reduce single expression:

x % 2l <> 0l && seq.forall (fun -> x % <> 0l) { 3l .. x/2l } 

although see no reason treating 2 differently, can do:

let isprime x =      { 2l .. x/2l } |> seq.forall (fun -> x % <> 0l)