In this puzzle, letters have been consistently replaced by digits, where:
• The same letter is substituted by the same digit.
• Different letters are substituted by different digits.
• No number can contain any leading zero.
It is known that each of the numbers FOUR and FIVE is a perfect square.
Furthermore, it is known that the each of the numerical answers to the undernoted queries is constituted entirely by the digits 4 and 5.
• (a) How many squares are strictly between FOUR and FIVE?
• (b) What is one-tenth of TEN - FIVE + FOUR?
• (c) What is the lowest positive integer that must be added to TEN to make it a perfect square?
Determine the value of FORTUNE.
Source: Adapted from Enigma #674 that appeared on "New Scientist" in 1992.
I found one solution which satisfies all criteria
FOUR FIVE TEN (a) (b) (c) FORTUNE
1024 1369 895 4 55 5 1048259
--------------- summary of findings
I found three sets of two squares for FOUR and FIVE with the same first digit and all 6 other digits being distinct and also criterion (a), there are 4 or 5 squares between FOUR and FIVE. Below are [FOUR, FIVE], #of squares between FOUR and FIVE
[[[1024, 1369], 4], [[2304, 2916], 5], [[7056, 7921], 4]]
This narrowed a little when also required to satisfy condition b. The possible values for "TEN" were: 795, 895, 567, 568
And if "one-tenth of TEN - FIVE + FOUR" must contain only 4s and 5s, the possible values for this calculation: 45, 55, -4.5, -4.4
And the candidates for FORTUNE: 1047259, 1048259, 2345076, 2345086
Finally, for the condition in (c):
The possible triplets below are "TEN", # added to TEN to make a square, that square:
[795, 46, 841], [895, 5, 900], [567, 9, 576], [568, 8, 576]
the second number of each group is the number to be added, and this must contain only 4s and 5s, so all that is left is [895, 5, 900]
--------------- CODE
from itertools import combinations
lo = int(1000**.5)+1
hi = int(10000**.5)
squares4d = [i*i for i in range(lo,hi)]
squaresBy1st = [[] for i in range (10)]
non45digits = [str(i) for i in range (10) if i !=4 and i !=5]
fourfives = []
squaresbetween = []
criterion_a = []
criterion_b = []
possible_ten_values = []
poss_one_tenth_of_mysum_values = []
values_for_FORTUNE = []
mustbeaddedtoTEN = []
for s in squares4d:
firstDigit = int(str(s)[0])
squaresBy1st[firstDigit].append(s)
for fD in range(1,10):
for pair in combinations(squaresBy1st[fD],2):
uniqueDigits = len( set(str(pair[0])+str(pair[1]) ))
if uniqueDigits == 7:
fourfives.append(sorted(list(pair)))
for ff in fourfives:
betw = squares4d.index(ff[1]) - squares4d.index(ff[0]) - 1
if ( betw ) == 4:
criterion_a.append(ff)
squaresbetween.append([ff,betw])
if ( betw ) == 5:
criterion_a.append(ff)
squaresbetween.append([ff,betw])
print('Meets criterion a', squaresbetween)
allDigits = [i for i in range(10)]
for p in criterion_a:
four = str(p[0])
five = str(p[1])
f = int(four[0])
o = int(four[1])
u = int(four[2])
r = int(four[3])
i = int(five[1])
v = int(five[2])
e = int(five[3])
takens = [f,o,u,r,i,v,e]
pool = [i for i in allDigits if i not in takens]
for t in pool:
for n in pool:
if n == t:
continue
ten = 100*t + 10*e + n
mysum = ten - int(five) + int(four)
quotient = mysum/10
if quotient%1 == 0:
quotient = int(quotient)
all4sand5s = True
for c in non45digits:
if c in str(quotient):
all4sand5s = False
count10 = 0
if all4sand5s:
fortune = f*10**6 + o*10**5 + r*10**4 + t*10**3 + u*10**2 + n*10**1 + e
values_for_FORTUNE.append(fortune)
possible_ten_values.append(ten)
poss_one_tenth_of_mysum_values.append(quotient)
if p not in criterion_b:
criterion_b.append(p)
squares = [i*i for i in range(1000)]
for item in possible_ten_values:
largerSquares = [i for i in squares if i > item]
myDifference = sorted(largerSquares)[0] - item
mustbeaddedtoTEN.append([item, myDifference, item + myDifference])
print('Meets criteria a and b', criterion_b)
print('one tenth of sum', poss_one_tenth_of_mysum_values)
print(possible_ten_values)
print(values_for_FORTUNE)
print(mustbeaddedtoTEN)
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Posted by Larry
on 2023-02-03 10:46:35 |
Solution
