S2T2

Jim Randell 8:35 am on 17 July 2019 Permalink Reply
Tags: by: Stephen Hogg ( 62 )   

• 

  Jim Randell 8:35 am on 17 July 2019 Permalink | Reply

  This Python program collects the possible multiplications where a reversed answer would be correct in a non-decimal base. It then uses the [[ filter_unique() ]] function (from the enigma.py library) to find situations where the answer for a specific digit indicate a unique multiplication question.

  This Python program runs in 60ms. (Internal runtime is 673µs).

  Run: [ @replit ]

  from enigma import (subsets, irange, multiply, div, unpack, filter_unique, printf)
  
  # collect possible questions
  ss = list()
  
  # consider triples composed of single digits (i, j, k)
  for t in subsets(irange(1, 9), size=3, select='R'):
    n = multiply(t)
    if not (9 < n < 100): continue
  
    # n = 10x + y = by + x
    (x, y) = divmod(n, 10)
    if x == y or y == 0: continue
    b = div(n - x, y)
    if b is None: continue
    if not (x < b and y < b and b != 10): continue
  
    # record tuples (<triple>, <product>, <tens digit>, <units digit>, <base>)
    printf("[{n} (base 10) = {y}{x} (base {b}), {n} -> {t}]")
    ss.append((t, n, x, y, b))
  
  # K asks if one of the numbers in the triple is a particular value k
  for k in irange(1, 9):
    # K asks if the digit k is (at least) one of the digits in the question
    # and from the answer can determine both the question and J's answer
    f = unpack(lambda t, n, x, y, b: k in t)
    g = unpack(lambda t, n, x, y, b: (t, n))
    rs = filter_unique(ss, f, g).unique
    # output unique solutions
    for r in rs:
      (t, n, x, y, b) = r
      printf("[k={k} -> {z}] question: {t[0]}x{t[1]}x{t[2]} (= {n}), jay's answer: {y}{x} (in base {b} = {n})", z=f(r))
  

  Solution: Jay’s answer was 48.

  It is odd that the setter asks for Jay’s written answer, which can refer to multiple possible written questions, rather than the three digits of the written question, which are unique.

  There are 9 possible multiplications where a reversed answer would be correct in a non-decimal base:

  1×3×7 = 21 [base 10] = 12 [base 19]
  1×6×7 = 2×3×7 = 42 [base 10] = 24 [base 19]
  1×7×9 = 3×3×7 = 63 [base 10] = 36 [base 19]
  2×6×7 = 3×4×7 = 84 [base 10] = 48 [base 19]
  1×9×9 = 3×3×9 = 81 [base 10] = 18 [base 73]

  Kay asks if the question includes the digit 4, and receives the answer “Yes”, this narrows the question down to a single possibility:

  3×4×7 = 84 [base 10] = 48 [base 19]

  This is the only digit/answer combination which gives a unique answer for the question.

  However, if Kay had asked about the digit 7, and received a “No” answer, there would have been two possible questions, but he would still be able to work out Jay’s answer:

  1×9×9 = 3×3×9 = 81 [base 10] = 18 [base 73]

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• 

  GeoffR 8:18 pm on 23 July 2019 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  % triplets of single value positive integers
  var 1..9: i; var 1..9: j; var 1..9: k;
  
  % the correct two digit result in base 10
  var 9..99: nbr;    
  
  % tens and units digits of the result                    
  var 1..9: t; var 1..9: u;  
  
  % the number base for reversed digits
  var 11..99: b;              
  
  % order triplet digits
  constraint i <= j /\ j <= k;
  
  constraint nbr = i * j * k;
  
  % find tens and units digits of 2-digit number
  constraint t = nbr div 10 /\ u == nbr mod 10;
  
  % number is correct in base b if digits are reversed
  constraint nbr == 10*t + u /\ nbr == b*u + t;
  
  solve satisfy;
  
  output [show(i) ++ " * " ++ show(j) ++ " * " ++ show(k) ++ " == " 
  ++show(nbr) ++ " (base 10) == " ++ show(u) ++ show(t) ++ 
  " (base " ++ show(b) ++ ")"]
              
  % 1 * 3 * 7 == 21 (base 10) == 12 (base 19)
  % 3 * 3 * 9 == 81 (base 10) == 18 (base 73)
  % 1 * 9 * 9 == 81 (base 10) == 18 (base 73)
  % 1 * 6 * 7 == 42 (base 10) == 24 (base 19)
  % 2 * 3 * 7 == 42 (base 10) == 24 (base 19)
  % 3 * 3 * 7 == 63 (base 10) == 36 (base 19)
  % 1 * 7 * 9 == 63 (base 10) == 36 (base 19)
  
  % Jay's answer - digit 4 gives a unique result  
  % 3 * 4 * 7 == 84 (base 10) == 48 (base 19)  <<<
  
  % 2 * 6 * 7 == 84 (base 10) == 48 (base 19)
  % ----------
  % ==========         
  

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