S2T2

Jim Randell 9:44 am on 1 August 2024 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

• 

  Jim Randell 9:44 am on 1 August 2024 Permalink | Reply

  This Python program runs in 81ms. (Internal runtime is 381µs).

  Run: [ @replit ]

  from enigma import (
    powers, is_prime, filter_unique, dsum,
    is_duplicate, singleton, group, printf
  )
  
  # find 4-digit squares, where the first 3 digits form a prime
  sqs = list(n for n in powers(32, 99, 2) if is_prime(n // 10))
  
  # select numbers unique by digit sum
  ns = filter_unique(sqs, f=dsum).unique
  
  # group results by whether there are repeated digits
  g = group(ns, by=is_duplicate)
  
  # look for unique values
  for (k, vs) in g.items():
    n = singleton(vs)
    if n is not None:
      printf("(duplicate = {k}) => n = {n}")
  

  Solution: Callum’s number was: 5476.

  There are 8 candidate squares that can be truncated to a prime, but only 3 have a unique digit sum:

  dsum = 10 → 2116
  dsum = 13 → 3136
  dsum = 19 → 1936, 4096
  dsum = 22 → 5476
  dsum = 25 → 5776, 7396, 8836

  Two of these 3 have repeated digits, so Callum must have answered that square did not have repeated digits, which uniquely identifies 5476 as his number.

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• 

  GeoffR 4:17 pm on 1 August 2024 Permalink | Reply

  
  from collections import defaultdict
  from enigma import is_prime, dsum
  
  ans = defaultdict(list)
  
  sq4 = [ x**2 for x in range(32, 100)]
  
  for n1 in sq4:
      flag = 0  # for repeating or non-repeating digits
      n2 = int(str(n1)[0:3])
      if is_prime(n2):
          # non rpeeating digits in squares
          if len(set(str(n1))) == len(str(n1)):
              flag = 1
              ans[(flag, dsum(n1))] += [n1]
          # repeating digits in squares
          if len(set(str(n1))) != len(str(n1)):
              flag = 2
              ans[(flag, dsum(n1))] += [n1]
  
  for k, v in ans.items():
      print(k,v)
  
  '''
  (1, 19) [1936, 4096] <<< multiple squares for digit sum
  (2, 10) [2116] <<< same number of repeating digits as square 3136
  (2, 13) [3136] <<< same number of repeating digits as square 2116
  (1, 22) [5476]  <<< Answer
  (2, 25) [5776, 8836] <<< multiple squares for digit sum
  (1, 25) [7396]  <<< same digit sum as squares 5776, 8836
  '''
  
  
  
  

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• 

  Ruud 8:50 pm on 1 August 2024 Permalink | Reply

  from istr import istr
  import math
  import collections
  
  
  def is_prime(n):
      if n < 2:
          return False
      if n % 2 == 0:
          return n == 2  # return False
      k = 3
      while k * k <= n:
          if n % k == 0:
              return False
          k += 2
      return True
  
  
  squares_per_sum_digits = collections.defaultdict(list)
  
  for i in range(round(math.sqrt(1000)), round(math.sqrt(10000))):
      square = istr(i * i)
      if is_prime(square[:-1]):
          prime = square[:-1]
          sum_digits = sum(n for n in square)
          squares_per_sum_digits[sum_digits].append(square)
  
  potential_squares = [squares[0] for squares in squares_per_sum_digits.values() if len(squares) == 1]
  has_repeated_digits = collections.defaultdict(list)
  for square in potential_squares:
      has_repeated_digits[len(set(square))!=4].append(square)
  
  solution = [square[0] for square in has_repeated_digits.values() if len(square) == 1]
  print(*solution)
  

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• 

  GeoffR 7:57 pm on 3 August 2024 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:A; var 0..9:B; var 1..9:C; var 0..9:D;
  var 1024..9081: sq_dig4 == 1000*A + 100*B + 10*C + D;
  
  set of int: sq4 = {n * n | n in 32..99};  
  
  predicate is_prime(var int: x) = 
  x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x)))))
   ((i < x) -> (x mod i > 0));
   
  % Square with last digit deleted is a prime
  constraint is_prime(100*A + 10*B + C);
  constraint sq_dig4 in sq4;
  
  solve satisfy;
  
  output[ "Digit sum = " ++ show(A + B + C + D) ++
  ", Square = " ++ show(sq_dig4) ];
  
  % Digit sum = 10, Square = 2116 - repeating digits - 2nd stage elimination
  % ----------
  % Digit sum = 19, Square = 1936 - not a square with a single digit sum - 1st stage elimination
  % ----------
  % Digit sum = 13, Square = 3136 - repeating digits - 2nd stage eliminations
  % ----------
  % Digit sum = 25, Square = 8836 - not a square with a single digit sum - 1st stage elimination
  % ----------
  % Digit sum = 22, Square = 5476 - *** ANSWER *** (unique digit sum, no repeating digits)
  % ----------
  % Digit sum = 25, Square = 5776 - not a square with a single digit sum - 1st stage elimination
  % ----------
  % Digit sum = 19, Square = 4096 - not a square with a single digit sum - 1st stage elimination
  % ----------
  % Digit sum = 25, Square = 7396 - not a square with a single digit sum - 1st stage elimination
  % ----------
  % ==========
  

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