New Scientist Enigma 1000 – One Thousand Times
by Richard England
From Issue #2155, 10th October 1998
Since M is the Roman numeral for 1000, we can say that with this puzzle New Scientist has published its Enigma M times – which is significant because:
ENIGMA ÷ M = TIMES
In this problem each letter stands for a different digit, and the same letter represents the same digit wherever it appears. No number starts with a zero.
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Brian Gladman permalink12345678910111213from itertools import permutationsfor p in permutations('0123456789', 5):T, I, M, E, S = pif '0' not in {M, T}:TIMES = ''.join(p)ENIGMA = str(int(M) * int(TIMES))if len(ENIGMA) == 6:e, N, i, G, m, A = ENIGMAif (e, i, m) == (E, I, M) and len(set(p).union([N, G, A])) == 8:print(f"{ENIGMA} / {M} == {TIMES}")123456789from alphasum import AlphaSumfor m in range(1, 10):c = AlphaSum(['TIMES'] * m, 'ENIGMA')for d in c.solve():if m == d['M']:print(c.output())
