Enigmatic Code
Programming Enigma Puzzles
Enigma 589: PG remaining
Posted by on 23 November 2020
From New Scientist #1742, 10th November 1990 [link]
In this division sum, each letter stands for a different digit. Rewrite the sum with the letters replaced by digits.
[enigma589]
Enigmatic Code 
We can use the [[
SubstitutedDivision]] solver from the enigma.py library to solve this puzzle.The following run file executes in 92ms.
Run: [ @repl.it ]
Solution: The correct sum is: 80142 ÷ 73 = 1097 (remainder 61).
from itertools import permutations # By the shape of the sum, g = 1 and e = 0 g, e = str(1), str(0) for p1 in permutations('23456789'): d, s, m, b, a, p, r, x = p1 ds, pg = int(d + s), int(p + g) gerd = int(g + e + r + d) megba = int(m + e + g + b + a) # main equation if ds * gerd + pg == megba: #intermediate sums pxd, dgb = int(p + x + d), int(d + g + b) xd, xda, xgg = int(x + d), int(x + d + a), int(x + g + g) d, r = int(d), int(r) if r * ds == pxd and dgb - pxd == xd: if d * ds == xgg and xda - xgg == pg: print(f"Sum is {megba} / {ds} = {gerd}, remainder {pg}") # Sum is 80142 / 73 = 1097, remainder 61