Three Wise Men

December 28, 2012

We have a simple Christmas puzzle today:

Three wise men named Caspar, Melchior and Balthazar went to Macy’s Department Store on 34th Street to buy gifts of gold, frankincense, and myrrh. When they took their gifts to the cashier, she multiplied the prices of the items and found a total price of $65.52, but the wise men noticed that she multiplied the numbers and asked her to compute the total price again, but this time using addition instead of multiplication. When she did, the cashier found that the total was exactly the same. What were the individual prices of the three gifts?

Your task is to solve the puzzle and find the three prices. 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

Posted by programmingpraxis
Filed in Exercises
11 Comments »

11 Responses to “Three Wise Men”

  1. Paul said
    December 28, 2012 at 10:22 AM

    A Python version.

    def three_wise_men(sumabc):
    prodabc = sumabc * 10000
    for a in xrange(1,sumabc // 3):
        if prodabc % a == 0:
            prodbc = prodabc // a
            sumbc =sumabc - a
            for b in xrange(a, sumbc // 2):
                if prodbc % b == 0:
                    c = prodbc // b
                    if b + c == sumbc:
                        return a, b, c
            
print three_wise_men(6552)
  2. Programming Praxis – Three Wise Men « Bonsai Code said
    December 28, 2012 at 11:39 AM

    […] today’s Programming Praxis exercise, our goal is to solve a puzzle in which both the addition and […]

  3. Remco Niemeijer said
    December 28, 2012 at 11:40 AM

    My Haskell solution (see http://bonsaicode.wordpress.com/2012/12/28/programming-praxis-three-wise-men/ for a version with comments):

    main :: IO ()
main = print $ head [(a,b,c) | a <- [1..6550], b <- [1..a], let c = 6552 - a - b, a * b * c == 65520000]
  5. Mark Henderson said
    December 28, 2012 at 4:12 PM

    Takes a few minutes:

    def three_wise():
    r = [x for x in range(1, 6553)]
    return next((a, b, c) for a in r for b in r for c in reversed(r)
            if a + b + c == 6552
            if a * b * c == 65520000)

print three_wise()
  6. Overpants said
    December 28, 2012 at 10:02 PM

    This is an easier version of an earlier problem (https://programmingpraxis.com/2009/11/27/7-11/). Unlike the other problem, it seems that you can just brute force this one and still have it run fairly quickly.

  7. cosmin said
    December 29, 2012 at 6:19 AM

    Another Python solution:

    from math import sqrt, ceil

def divisors(n):
	f = [i for d in xrange(1, 1 + int(sqrt(n))) if n%d == 0 for i in [d, n/d]]
	if len(f) >= 2 and f[-1] == f[-2]: f.pop()
	return f

def three_wise_men(x):
	s, p = int(ceil(x*100)), int(ceil(x*1000000))
	return [(a, b, c) for a in divisors(p) if a <= s/3 for b in divisors(p/a) for c in [p/a/b] if a<=b<=c and a+b+c == s]

print three_wise_men(65.52)
  10. Heather said
    December 31, 2012 at 3:33 AM

    My fastest Ruby solution:

    $ time ruby wisemen.rb
    [0.52, 2.0, 63.0]

    real 0m1.919s
    user 0m1.795s
    sys 0m0.120s

    dollars = 65.52
cents = (dollars * 100).to_i

trips = []

i = 1
j = 1
halfcents = cents / 2
while true
  k = cents - i - j
  if k < j
    i += 1
    j = i
    if (i >= halfcents) || (k < 2)
      break
    else
      next
    end
  end
  trips << [i, j, k]
  j += 1
end

trips.each do |trip|
  trip = trip.map { |t| t / 100.0 }
  if trip.inject(:*) == dollars
    puts trip.inspect
    break
  end
end

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