Teaser 3026: Party time
From The Sunday Times, 20th September 2020 [link]
A four-digit number with different positive digits and with the number represented by its last two digits a multiple of the number represented by its first two digits, is called a PAR.
A pair of PARs is a PARTY if no digit is repeated and each PAR is a multiple of the missing positive digit.
I wrote down a PAR and challenged Sam to use it to make a PARTY. He was successful.
I then challenged Beth to use my PAR and the digits in Sam’s PAR to make a different PARTY. She too was successful.
What was my PAR?
[teaser3026]
GeoffR, 
Will Harris, and 1 other are discussing. 
S2T2 
Jim Randell 5:30 pm on 18 September 2020 Permalink |
This Python program uses the [[
SubstitutedExpression()]] general alphametic solver from the enigma.py library to find possible PARTYs. It then collects pairs of PARs together and looks for a PAR that has two possible pairings that share the same digits.It runs in 56ms.
Run: [ @repl.it ]
from collections import defaultdict from enigma import SubstitutedExpression, irange, subsets, nsplit, printf # find possible PARTYs p = SubstitutedExpression( ["CD % AB = 0", "ABCD % X = 0", "GH % EF = 0", "EFGH % X = 0", "ABCD < EFGH"], digits=irange(1, 9), answer="(ABCD, EFGH, X)", ) # collect results by PARs d = defaultdict(list) for (_, (ABCD, EFGH, X)) in p.solve(verbose=0): d[ABCD].append(EFGH) d[EFGH].append(ABCD) # collect the digits of a number digits = lambda n: sorted(nsplit(n)) # choose the initial PAR for (k, vs) in d.items(): # choose two related PARs ... for (p1, p2) in subsets(vs, size=2): # ... that have the same digits if digits(p1) == digits(p2): printf("{k} -> {p1}, {p2}")Solution: Your PAR was 1785.
1785 can be paired with both 2496 and 4692 (missing digit = 3) to make a PARTY.
There are 7 possible PARTYs:
There are only 2 PARs that appear in more than one PARTY: 1785 and 2754. Each of these appears in two PARTYs, but only 1785 is paired with two other PARs that share the same digits.
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Jim Randell 12:28 pm on 20 September 2020 Permalink |
Using the [[
SubstitutedExpression]] general alphametic solver from the enigma.py library.This run file executes in 94ms.
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Frits 9:12 pm on 19 September 2020 Permalink |
from enigma import nconcat from itertools import permutations, combinations difference = lambda a, b: [x for x in a if x not in b] def getPARs(ds, k): li = [] # check all permutations of digits for p1 in permutations(ds): # first digit can't be high if p1[0] > 4: continue n12 = nconcat(p1[:2]) n34 = nconcat(p1[2:]) n = 100 * n12 + n34 # number last two digits is a multiple of number first 2 digits # and four-digit number is a multiple of k if n34 % n12 != 0 or n % k != 0: continue # store the PAR li.append(n) return li sol = [] # loop over missing digits for k in range(1,10): digits = difference(range(1,10), [k]) # choose the lowest remaining digit low = min(digits) digits = difference(digits, [low]) # pick 3 digits to complement low for c1 in combinations (digits, 3): par1 = getPARs((low,) + c1, k) if par1: par2 = getPARs(difference(digits, c1), k) # PAR(s) related to two or more PARs? if len(par1) == 1 and len(par2) == 1: continue if len(par2) == 1: sol.append([par2, par1]) elif par2: sol.append([par1, par2]) for i in range(len(sol)): print(f"{sol[i][0]} --> {sol[i][1]}")LikeLike
Will Harris 3:14 pm on 19 December 2020 Permalink |
Not much help now but wrote this for my coursework project using GNU prolog [ https://www.tutorialspoint.com/execute_prolog_online.php ]
:- initialization(main). % This program has been developed for GNU Prolog | https://www.tutorialspoint.com/execute_prolog_online.php %Question 2.1 convert(P, Q) :- Q is P - 48, Q > 0. unique(List) :- sort(List, Sorted), length(List, OriginalLength), length(Sorted, SortedLength), OriginalLength =:= SortedLength. multiple_of([A,B,C,D]) :- get_num(A,B,X), get_num(C,D,Y), check_mod(X,Y). get_num(P,Q,R) :- R is P * 10 + Q. check_mod(N,M) :- M mod N =:= 0. par(Number) :- number_codes(Number,ListCharCodes), maplist(convert,ListCharCodes,ListDigits), length(ListDigits, 4), unique(ListDigits), multiple_of(ListDigits). %Question 2.2 candidate_pars(P, Q, []) :- P > Q. candidate_pars(P, Q, [P|W]) :- par(P), P1 is P + 1, candidate_pars(P1, Q, W). candidate_pars(P, Q, W) :- P1 is P + 1, candidate_pars(P1, Q, W). pars( PARS ) :- candidate_pars(1234, 9876, PARS). %Question 2.3 missing_digit(V,W,CombDigits) :- number_codes(V, ListOne), number_codes(W, ListTwo), maplist(convert, ListOne, ListDigitsOne), maplist(convert, ListTwo, ListDigitsTwo), append(ListDigitsOne, ListDigitsTwo, CombDigits). party(V, W) :- missing_digit(V, W, CombDigits), subtract([1,2,3,4,5,6,7,8,9], CombDigits, MissingDigits), length(MissingDigits, 1), par(V), par(W), nth(1, MissingDigits, M), check_mod(M,V), check_mod(M,W). %Question 2.4 comparison(PARS, I, J) :- member(I, PARS), member(J, PARS), party(I,J). partys( PARTYS ) :- pars(PARS), findall([I,J], comparison(PARS, I, J), PARTYS). main :- partys( PARTYS ), write(PARTYS). /* There is only a single solution to Teaser 3026: There are only 2 PARs that appear in more than one PARTY: 1785 and 2754. Each of these appears in two PARTYs, but only 1785 is paired with two other PARs that share the same digits. 1785 can be paired with both 2496 and 4692 (missing digit = 3) to make a PARTY. */LikeLike
Jim Randell 3:26 pm on 19 December 2020 Permalink |
@Will: Thanks for your comment.
I’m not sure if the code came through correctly (WordPress can get confused by text with < and > characters in it).
If it’s wrong, can you post again with the code in:
and I’ll fix it up.
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GeoffR 4:57 pm on 16 December 2021 Permalink |
from itertools import permutations digits = set('123456789') for p1 in permutations(digits, 5): A, B, C, D, I = p1 # my PAR is ABCD and I is the missing digit ABCD = int(A + B + C + D) if ABCD % int(I) !=0: continue AB, CD = int(A + B), int(C + D) if not(CD % AB == 0):continue # find another two PARs to make two PARTYs q1 = digits.difference(p1) for p2 in permutations(q1): E, F, G, H = p2 EFGH = int(E + F + G + H) if not EFGH % int(I) == 0: continue EF, GH = int(E + F), int(G + H) if not (GH % EF == 0):continue # form 2nd PAR with digits (E, F, G, H) for p3 in permutations(p2): e, f, g, h = p3 efgh = int(e + f + g + h) # there must be two different PARs from EFGH if EFGH == efgh:continue if efgh % int(I) != 0: continue ef, gh = int(e + f), int(g + h) if not (gh % ef == 0):continue print(f"My PAR = {ABCD}") print(f"Sam / Beth's PARTY = {ABCD}, {EFGH}") print(f"Sam / Beth's PARTY = {ABCD}, {efgh}") print() # My PAR = 1785 # Sam / Beth's PARTY = 1785, 4692 # Sam / Beth's PARTY = 1785, 2496LikeLike