Teaser 2898: My valentine
From The Sunday Times, 8th April 2018 [link] [link]
Using only positive digits, MILLY wrote down a two-digit number and a nine-digit number, both numbers being divisible by 14.
Splitting her nine-digit number into three three-digit numbers, she noticed that all three numbers were also divisible by 14.
Systematically replacing different digits by different letters, her two-digit number and nine-digit number ended up as MY VALENTINE.
What number is represented by MILLY?
[teaser2898]
GeoffR, 
Frits, and 
Lise Andreasen and 
John Crabtree and 
S2T2 
Jim Randell 10:06 am on 19 July 2019 Permalink |
This puzzle can be solved directly using the [[
SubstitutedExpression()]] solver from the enigma.py library.The following run file executes in 131ms.
Run: [ @repl.it ]
Solution: MILLY = 17224.
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GeoffR 10:46 am on 20 July 2019 Permalink |
Gave up using Geocode Solver after three minutes but Chuffed Solver produced a solution in a reasonable time.
@Jim: Any idea why solvers run-times can vary so much?
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Jim Randell 11:10 am on 23 July 2019 Permalink |
@GeoffR: Using integer literals for the powers of 10 instead of [[
pow()]] seems to make the model acceptable to the [[gecode]] solver.LikeLike
GeoffR 12:20 pm on 23 July 2019 Permalink |
@Jim: Thanks for the reply, yes that suggestion now works OK for the Geoode solver. Perhaps the Chuffed solver works better for powers as it looks as though it is based on C++
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Jim Randell 1:16 pm on 23 July 2019 Permalink |
I suspect that one of the solvers notices that they are constant terms and calculates them once, but the other one doesn’t and keeps re-evaluating them.
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Frits 11:15 am on 18 September 2020 Permalink |
# concatenate list of integers to_num = lambda *args: int("".join(map(str, args))) # make sure loop variable value is not equal to previous ones def domain(v): # find already used loop values ... vals = set() # ... by accessing previously set loop variable names for s in lvd[v]: vals.add(globals()[s]) return [x for x in range(1,10) if x not in vals] # set up dictionary of for-loop variables lv = ['M', 'Y', 'E', 'N', 'T', 'I', 'V', 'A', 'L'] lvd = {v: lv[:i] for i, v in enumerate(lv)} sol = set() # Solve puzzle by testing all posssible values for M in range(1,10): for Y in domain('Y'): if (to_num(M, Y) % 14 != 0): continue for E in domain('E'): for N in domain('N'): for T in domain('T'): if (to_num(E, N, T) % 14 != 0): continue for I in domain('I'): if (to_num(I, N, E) % 14 != 0): continue for V in domain('V'): for A in domain('A'): for L in domain('L'): if (to_num(V, A, L) % 14 != 0): continue sol.add(to_num(M, I, L, L, Y)) for s in sol: print(s) # Output: # # 17224LikeLike
Frits 11:35 am on 18 September 2020 Permalink |
As VAL000000 is divisible by 14, ENT000 is divisible by 14 and INE is divisible by 14 we don’t need to verify that VALENTINE is divisible by 14.
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Frits 2:31 pm on 30 November 2021 Permalink |
Using choose().
from enigma import choose # concatenate list of integers to_num = lambda *args: int("".join(map(str, args))) S = list(choose(range(1, 10), [None, (lambda M, Y: Y not in {M} and to_num(M, Y) % 14 == 0), (lambda M, Y, E: E not in {M, Y}), (lambda M, Y, E, N: N not in {M, Y, E}), (lambda M, Y, E, N, T: T not in {M, Y, E, N} and to_num(E, N, T) % 14 == 0), (lambda M, Y, E, N, T, I: I not in {M, Y, E, N, T} and to_num(I, N, E) % 14 == 0), (lambda M, Y, E, N, T, I, V: V not in {M, Y, E, N, T, I}), (lambda M, Y, E, N, T, I, V, A: A not in {M, Y, E, N, T, I, V}), (lambda M, Y, E, N, T, I, V, A, L: L not in {M, Y, E, N, T, I, V, A} and to_num(V, A, L) % 14 == 0 and to_num(V, A, L, E, N, T, I, N, E) % 14 == 0), ])) print("MILLY:", *set("".join(str(x) for x in [s[0], s[5], s[8], s[8], s[1]]) for s in S))LikeLike
Frits 4:28 pm on 30 November 2021 Permalink |
With list comprehensions.
# even digits are reserved for Y, T, E and L digits = set("123456789") odds = set("13579") evens = digits.difference(odds) # even digits are reserved for Y, T, E and L # so M must be odd MY = set(s for x in range(10, 100) if x % 14 == 0 and len(set((s := str(x)))) == 2 and s[0] in odds and '0' not in s) # list of 3 digit numbers divisible by 14 # middle digit (A or N) must be odd dgts3 = [s for x in range(100, 1000) if x % 14 == 0 and len(set((s := str(x)))) == 3 and s[1] in odds and '0' not in s] ooe = [x for x in dgts3 if x[0] in odds] # odd, odd, even # ENT and INE are both divisible by 14 # N and I must be odd ENT_INE = [(x, y) for x in dgts3 for y in ooe if y[2] + y[1] == x[:2] and y[0] != x[2]] ENT_INE_VAL = [(x, y, z) for x, y in ENT_INE for z in ooe if "".join(sorted(digits.difference(x + y + z))) in MY] sol = set() for ENT, INE, VAL in ENT_INE_VAL: t = ENT + INE + VAL M = odds.difference([x for x in t if x in odds]) Y = evens.difference([x for x in t if x in evens]) sol.add(M.pop() + t[3] + t[8] + t[8] + Y.pop()) for s in sol: print("MILLY:", s)LikeLike
GeoffR 11:39 am on 1 December 2021 Permalink |
from itertools import permutations for p1 in permutations('123456789', 2): M, Y = p1 MY = int(M + Y) if MY % 14 != 0: continue q1 = set('123456789').difference([M, Y]) # 1st part(VAL) of VALENTINE for p2 in permutations(q1, 3): V, A, L = p2 VAL = int(V + A + L) if VAL % 14 != 0:continue q2 = q1.difference([V, A, L]) # 2nd part(ENT) of VALENTINE for p3 in permutations(q2, 4): E, N, T, I = p3 ENT = int(E + N + T) if ENT % 14 != 0:continue # 3rd part(INE) of VALENTINE INE = int(I + N + E) if INE % 14 != 0: continue VALENTINE = int(V + A + L + E + N + T + I + N + E) # Stated condition, but not necessary #if VALENTINE % 14 != 0:continue MILLY = int(M + I + L + L + Y) print(f"MY={MY}, VALENTINE={VALENTINE}, MILLY={MILLY}") # MY=14, VALENTINE=392658756, MILLY=17224 # MY=14, VALENTINE=532896798, MILLY=17224LikeLike