Enigmatic Code
Programming Enigma Puzzles
Enigma 1506: China games
Posted by on 9 October 2012
From New Scientist #2668, 9th August 2008
In the three arithmetic additions below, the number 2008 is given and the undisclosed digits have been replaced by letters and smiley faces. Different letters stand for different digits (the same letter consistently stands for the same digit), each face can be any digit, and leading digits cannot be zero.
Find the value of GAMES.
[enigma1506]
Enigmatic Code 
The following Python program runs in 80ms.
from itertools import permutations from enigma import irange, nconcat, printf # possible digits ds0 = set(irange(0, 9)) # from the first sum it follows (ES + IN + NA + IN) % 100 = 8 for (A, E, I, N, S) in permutations(ds0, 5): if (10 * E + S + 20 * I + 12 * N + A) % 100 != 8: continue # try the remaining digits ds1 = ds0.difference((A, E, I, N, S)) for (C, G, H, M) in permutations(ds1, 4): # no leading zeros if 0 in (C, G, I): continue # from sum 3 it follows C + G + I + 1 + (2 + H + A + N)/10 < 10 if C + G + I + (2 + H + A + N) // 10 > 8: continue GAMES = nconcat(G, A, M, E, S) IN = nconcat(I, N) CHINA = nconcat(C, H, I, N, A) # sum 2: # a2008 + GAMES + INbcd = CHINA a0bcd = CHINA - (2008 + GAMES + IN * 1000) if not(9999 < a0bcd < 100000): continue s = str(a0bcd) if s[1] != '0': continue # sum 3: # e2008 + CHINA + GAMES + INfgh = i2008 # (2008 + HINA + AMES + Nfgh) % 10000 = 2008 Nfgh = (2008 - (2008 + CHINA % 10000 + GAMES % 10000) % 10000) % 10000 (n, fgh) = divmod(Nfgh, 1000) if n != N: continue for e in irange(1, 9): i2008 = e * 10000 + 2008 + CHINA + GAMES + IN * 1000 + fgh (i, r) = divmod(i2008, 10000) if not(0 < i < 10 and r == 2008): continue # sum 1: # GAMES + xxxIN + CHINA + yyyIN = z2008 # where 99 < xxx, yyy < 1000 for z in irange(1, 9): (xy, r) = divmod(z * 10000 + 2008 - (GAMES + 2 * IN + CHINA), 100) if not(199 < xy < 1999 and r == 0): continue printf("[1] {GAMES} + xxx{IN} + {CHINA} + yyy{IN} = {z}2008, xxx + yyy = {xy}") printf("[2] {a}2008 + {GAMES} + {IN}{bcd} = {CHINA}", a=s[0], bcd=s[2:]) printf("[3] {e}2008 + {CHINA} + {GAMES} + {IN}{fgh} = {i}2008") printf("GAMES={GAMES} IN={IN} CHINA={CHINA}")Solution: GAMES = 23785.
% A Solution in MiniZinc % Lower case letters represent smiley faces % GAMES h2008 m2008 % abcIN GAMES CHINA % CHINA INijk GAMES % defIN ----- INnpq % ----- CHINA ----- % g2008 ----- r2008 % ----- ----- include "globals.mzn"; var 0..9:G; var 0..9:A; var 0..9:M; var 0..9:E; var 0..9:S; var 0..9:I; var 0..9:N; var 0..9:C; var 0..9:H; constraint all_different([G, A, M, E, S, I, N, C, H]); var 0..9:a; var 0..9:b; var 0..9:c; var 0..9:d; var 0..9:e; var 0..9:f; var 0..9:g; var 0..9:h; var 0..9:i; var 0..9:j; var 0..9:k; var 0..9:m; var 0..9:n; var 0..9:p; var 0..9:q; var 0..9:r; constraint G != 0 /\ a != 0 /\ C != 0 /\ d != 0 /\ h != 0 /\ I != 0 /\ m != 0 /\ r != 0; var 10000..99999:GAMES = 10000*G + 1000*A + 100*M + 10*E + S; var 10000..99999:abcIN = 10000*a + 1000*b + 100*c + 10*I + N; var 10000..99999:CHINA = 10000*C + 1000*H + 100*I + 10*N + A; var 10000..99999:defIN = 10000*d + 1000*e + 100*f + 10*I + N; var 10000..99999:g2008 = g*10000 + 2008; var 10000..99999:h2008 = h*10000 + 2008; var 10000..99999:INijk = 10000*I + 1000*N + 100*i + 10*j + k; var 10000..99999:m2008 = m*10000 + 2008; var 10000..99999:INnpq = 10000*I + 1000*N + 100*n + 10*p + q; var 10000..99999:r2008 = r*10000 + 2008; constraint GAMES + abcIN + CHINA + defIN = g2008; constraint h2008 + GAMES + INijk = CHINA; constraint m2008 + CHINA + GAMES + INnpq = r2008; solve satisfy; output [ "GAMES = " ++ show(GAMES) ++ "\n" ] ++ [ show(GAMES) ++ " + " ++ show(abcIN) ++ " + " ++ show(CHINA) ++ " + " ++ show(defIN) ++ " = " ++ show(g2008) ++ "\n"] ++ [ show(h2008) ++ " + " ++ show(GAMES) ++ " + " ++ show(INijk) ++ " = " ++ show(CHINA) ++ "\n"] ++ [ show(m2008) ++ " + " ++ show(CHINA) ++ " + " ++ show(GAMES) ++ " + " ++ show(INnpq) ++ " = " ++ show(r2008) ]; % GAMES = 23785 % 23785 + 11910 + 46103 + 10210 = 92008 % 12008 + 23785 + 10310 = 46103 % 12008 + 46103 + 23785 + 10112 = 92008 % ------------------ % Finished in 58msecAllowing for numbers containing smiley faces to take different values, I found 22 solutions to the three equations in multiple output configuration for MiniZinc.
However, in all case, GAMES = 23785, IN = 10, CHINA = 46103 and the totals of the first and third equations was 92008.