Enigmatic Code
Programming Enigma Puzzles
Enigma 774: Sting in the tail
Posted by on 29 April 2024
From New Scientist #1929, 11th June 1994 [link]
In the above division sum sixes and zeros have been inserted wherever they appear. X stands for any digit, excepting 6 and zero but not necessarily the same digit throughout.
Find the value of the divisor XX.
[enigma774]
Enigmatic Code 
If we multiply the dividend (66) by 10^8 we can treat this as an integer division, and use the [[
SubstitutedDivision]] solver from the enigma.py library to solve it.The following run file executes in 91ms. (Internal runtime of the generated program is 1.4ms).
Run: [ @replit ]
Solution: The divisor of the sum is 73.
The completed sum looks like this:
And we can reconstruct the sum programatically using the [[
output_div()]] function in the enigma.py library:from enigma import (SubstitutedDivision, output_div) p = SubstitutedDivision.from_file("enigma774.run") for s in p.solve(verbose=0): output_div(s.a, s.b, pre=" ", start="", end="")% python3 enigma774.py 90410958 ------------ 73 ) 6600000000 657 --- 300 292 --- 80 73 -- 700 657 --- 430 365 --- 650 584 --- 66 (rem) ===In fact there is only one possible divisor that gives a repetend with a matching shape:
from enigma import (irange, recurring, match, format_recurring, printf) # find recurring fractions with a matching repetend for d in irange(67, 99): (i, nr, rr) = recurring(66, d) if (not nr) and match(rr, "[!6]0[!6][!6]0[!6][!6][!6]"): printf("66/{d} = {x}", x=format_recurring((i, nr, rr)))from istr import istr from itertools import product x_values = "12345789" for divisor in istr.concat(product(x_values, repeat=2)): quotient = 6600000000 / divisor if all(j in (("0", x_values)[i == "x"]) for i, j in zip("x0xx0xxx", quotient)): print(divisor, quotient)