Generate random number of certain amount of digits

Step 1

First of all we need an extension of Int to generate a random number in a range.

extension Int {
    init(_ range: Range<Int> ) {
        let delta = range.startIndex < 0 ? abs(range.startIndex) : 0
        let min = UInt32(range.startIndex + delta)
        let max = UInt32(range.endIndex   + delta)
        self.init(Int(min + arc4random_uniform(max - min)) - delta)
    }
}

This can be used this way:

Int(0...9) // 4 or 1 or 1...
Int(10...99) // 90 or 33 or 11
Int(100...999) // 200 or 333 or 893

Step 2

Now we need a function that receive the number of digits requested, calculates the range of the random number and finally does invoke the new initializer of Int.

func random(digits:Int) -> Int {
    let min = Int(pow(Double(10), Double(digits-1))) - 1
    let max = Int(pow(Double(10), Double(digits))) - 1
    return Int(min...max)
}

Test

random(1) // 8
random(2) // 12
random(3) // 829
random(4) // 2374

Swift 5: Simple Solution

func random(digits:Int) -> String {
    var number = String()
    for _ in 1...digits {
       number += "\(Int.random(in: 1...9))"
    }
    return number
}

print(random(digits: 1)) //3
print(random(digits: 2)) //59
print(random(digits: 3)) //926

Note It will return value in String, if you need Int value then you can do like this

let number = Int(random(digits: 1)) ?? 0

Here is some pseudocode that should do what you want.

generateRandomNumber(20)
func generateRandomNumber(int numDigits){
   var place = 1
   var finalNumber = 0;
   for(int i = 0; i < numDigits; i++){
      place *= 10
      var randomNumber = arc4random_uniform(10)
      finalNumber += randomNumber * place
  }
  return finalNumber
}

Its pretty simple. You generate 20 random numbers, and multiply them by the respective tens, hundredths, thousands... place that they should be on. This way you will guarantee a number of the correct size, but will randomly generate the number that will be used in each place.

Update

As said in the comments you will most likely get an overflow exception with a number this long, so you'll have to be creative in how you'd like to store the number (String, ect...) but I merely wanted to show you a simple way to generate a number with a guaranteed digit length. Also, given the current code there is a small chance your leading number could be 0 so you should protect against that as well.