A Prime Number Puzzle

November 28, 2014

Today’s exercise comes from the world of recreational mathematics.

For each number n from 1 to 9 inclusive, find a number m that begins with the digit n, has n digits, has each two-digit sequence within m a different prime number, and is as small as possible.

When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

9 Responses to “A Prime Number Puzzle”

  1. Paul said

    In Python. Can be easily solved with greed.

    from collections import defaultdict
    
    primes2 = [11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73,
               79, 83, 89, 97]
    
    P = defaultdict(list)
    for p in primes2:
        s = str(p)
        P[s[0]].append(s)
    
    def solve(s):
        n = int(s)
        used = set()
        while len(s) < n:
            for nextp in P[s[-1]]:
                if nextp not in used:
                    used.add(nextp)
                    s += nextp[1]
                    break
        return int(s)
            
    for n in "123456789":
        print n, solve(n)
    
  2. Andras said

     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
    import scala.math._
    
    object PrimePuzzle {
      def sieve(s: Stream[Int]): Stream[Int] = s.head #:: sieve(s.tail.filter(_ % s.head != 0))
                                                      //> sieve: (s: Stream[Int])Stream[Int]
    
      val primes = sieve(Stream.from(2)).take(25).toList.map(n => n.toString).filter(s => s.length()==2)
                                                      //> primes  : List[String] = List(11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53
                                                      //| , 59, 61, 67, 71, 73, 79, 83, 89, 97)
                                                      
      
    
      def generatePrimePuzzles(n:Int, s:String, remainingPrimes:List[String]): Stream[String] = {
      	if (n < 2) "does not exist" #:: Stream.empty
      	else if (s.length()==n) Stream(s)
      	else {
      		val nextPrimes = remainingPrimes.filter(p=>p.head == s.last)
      		if(nextPrimes.nonEmpty)
      			generatePrimePuzzles(n, s+nextPrimes.head.tail, remainingPrimes.filterNot(p=>p==nextPrimes.head))
      		else
      			Stream.empty
      	}
      }                                               //> generatePrimePuzzles: (n: Int, s: String, remainingPrimes: List[String])Stre
                                                      //| am[String]
      
      for(n <- 1 to 9) yield (n, generatePrimePuzzles(n, n.toString, primes)(0))
                                                      //> res0: scala.collection.immutable.IndexedSeq[(Int, String)] = Vector((1,does 
                                                      //| not exist), (2,23), (3,311), (4,4113), (5,53113), (6,611317), (7,7113173), (
                                                      //| 8,83113717), (9,971131737))
    }
    
  3. Andras said

    Excuse me for the numbers, hilite.me does not format very well for pp…

  4. Andras said

    And it would be great if we could edit our comments.

  5. Francesco said

    Haskell:

    tenp = ["11","13","17","19","23","29","31","37","41","43","47","53","59",
    "61","67","71","73","79","83","89","97"]

    addN :: String -> Int -> [String] -> [String]
    addN s 0 _ = return s
    addN s i p = filter ((== last s) . head) p >>= \xc ->
    addN (s ++ [last xc]) (i-1) (filter (/= xc) p)

    main = print $ map (\n -> head $ addN (show n) (n-1) tenp) [1..9]

    I guess using >>= to generate partial permutations would lead to a shorter solution, but after a bit of fiddling I manage to do it.

  6. Francesco said
    tenp = ["11","13","17","19","23","29","31","37","41","43","47","53","59",
            "61","67","71","73","79","83","89","97"]
    
    
    addN :: String -> Int -> [String] -> [String]
    addN s 0 _ = return s
    addN s i p = let l = last s in                                             
                 filter ((==l) . head) p >>= \xc ->
                 addN (s ++ [last xc]) (i-1) (filter (/= xc) p)
    
    main = print $ map (\n -> head $ addN (show n) (n-1) tenp) [1..9]
    

    Once again, wrong format :s

  7. Josef Svenningsson said
    twoDigitPrimes = [11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
    
    puzzle n = minimum $ map mkNumber 
        [ m | nb <- sequence (replicate (n-1) [0..9])
            , let m = n:nb
            , allPairs (`elem` twoDigitPrimes) m ]
    
    allPairs p (x:y:xs) = p (mkNumber [x,y]) && allPairs p (y:xs)
    allPairs p _ = True
    
    mkNumber = foldl (\r x -> x + 10*r) 0
    
    main = putStrLn $ unlines $ [ show (puzzle n) | n <- [1..9] ]
    
  8. Mike said

    Once you write down all the 2-digit primes, it is clear that the problem is simple enough to do manually in less time than it would take to write a program. Nevertheless, here’s a recursive Python solution:

    ps = "11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97".split()
    
    def primepuzzle(n, m='', ps=ps, hd=[]):
    	if len(m) == n:
    		return int(m)
    
    	if not m:
    		return primepuzzle(n, str(n), ps, hd)
    
    	if ps[0][0] != m[-1]:
    		return primepuzzle(n, m, ps[1:], hd + ps[:1])
    	
    	return primepuzzle(n, m + ps[0][1], hd + ps[1:], [])
    
    for n in range(1,10):
    	print(n, primepuzzle(n))
    
    
  9. Claire said

    @Paul ,@Francesco Can you solve this problem in C++,please? Thank you.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: