From The Sunday Times, 21st January 1962 [link]
This unusual instrument is operated by selecting one of the four switch positions: A, B, C, D, and turning the power on. The effects are:
A: The pratching valve glows and the queech obulates;
B: The queech obulates and the urfer curls up, but the rumption does not get hot;
C: The sneeveling rod turns clockwise, the pratching valve glows and the queech fails to obulate;
D: The troglodyser gives off hydrogen but the urfer does not curl up.
Whenever the pratching valve glows, the rumption gets hot. Unless the sneeveling rod turns clockwise, the queech cannot obulate, but if the sneeveling rod is turning clockwise the troglodyser will not emit hydrogen. If the urfer does not curl up, you may be sure that the rumption is not getting hot.
In order to get milk chocolate from the machine, you must ensure:
(a) that the sneeveling rod is turning clockwise AND;
(b) that if the troglodyser is not emitting hydrogen, the queech is not obulating.
1. Which switch position would you select to get milk chocolate?
If, tiring of chocolate, you wish to receive the Third Programme, you must take care:
(a) that the rumption does not get hot AND;
(b) either that the urfer doesn’t curl and the queech doesn’t obulate or that the pratching valve glows and the troglodyser fails to emit hydrogen.
2. Which switch position gives you the Third Programme?
No setter was given for this puzzle.
This puzzle crops up in several places on the web. (Although maybe it’s just because it’s easy to search for: “the queech obulates” doesn’t show up in many unrelated pages).
And it is sometimes claimed it “appeared in a national newspaper in the 1930s” (although the BBC Third Programme was only broadcast from 1946 to 1967 (after which it became BBC Radio 3)), but the wording always seems to be the same as the wording in this puzzle, so it seems likely this is the original source (at least in this format).
“Omnibombulator” is also the title of a 1995 book by Dick King-Smith.
[teaser44]
Tony Brooke-Taylor, 
Frits, 
GeoffR, and 
John Crabtree and 
Hugh Casement and 
Tony Brooke-Taylor, and 
S2T2 
Jim Randell 4:52 pm on 11 December 2020 Permalink |
If the sum of the 5 terms of the geometric progression is t, and this is also the sum of the 5 term arithmetic progression (a, a + d, a + 2d, a + 3d, a + 4d), then we have:
So the sum must be a multiple of 5.
The following Python program runs in 44ms.
Run: [ @repl.it ]
from enigma import (irange, div, union, nsplit, is_square, printf) # generate geometric progressions with k elements # n = max value for 1st element, N = max value for all elements def geom(k, n=9, N=1000): # first element for a in irange(1, n): # second element is larger for b in irange(a + 1, N): # calculate remaining elements (gs, rs, x) = ([a, b], 0, b) while len(gs) < k: (x, r) = divmod(x * b, a) gs.append(x) rs += r if x > N: break if rs == 0: yield tuple(gs) # digit sum of a sequence dsums = lambda ns: sum(nsplit(n, fn=sum) for n in ns) # consider geometric progressions for G for gs in geom(5): x = div(sum(gs), 5) if x is None: continue # arithmetic progressions for M, with the same sum # and starting with a single digit for a in irange(1, 9): d = div(x - a, 2) if d is None: continue ms = tuple(irange(a, a + 4 * d, step=d)) ns = union([gs, ms]) if len(ns) != 10: continue # the sum of the digits in all the numbers is a perfect square r = is_square(dsums(ns)) if r is None: continue m = max(gs[-1], ms[-1]) printf("max = {m}; gs = {gs}, ms = {ms}; sum = {t}, dsum = {r}^2", t=5 * x)Solution: The highest numbered ticket is 80.
Note that the [[
geom()]] function will generate all possible geometric progressions, including those that have a non-integer common ratio, (e.g. for a 4 element progression we could have (8, 28, 48, 343)), but for a 5 element progression we would need the initial term to have a factor that is some (non-unity) integer raised to the 4th power, and the smallest possible value that would allow this is 2^4 = 16, which is not possible if the initial term is a single digit. So for this puzzle the solution can be found by considering only those progressions with an integer common ratio (which is probably what the puzzle wanted, but it could have said “integer multiple” to be completely unambiguous).LikeLike
GeoffR 7:48 pm on 11 December 2020 Permalink |
% A Solution in MiniZinc include "globals.mzn"; % terms of arithmetic series var 1..9: A1; var 2..100: A2;var 2..150: A3; var 2..250: A4; var 2..500: A5; % terms of geometric series var 1..9: G1; var 2..100: G2;var 2..150: G3; var 2..250: G4; var 2..500: G5; % geometric ratio and arithmetic difference var 2..9:Gratio; var 2..200:Adiff; % total sum of geometric and arithmetic series var 2..500: Gsum; var 2..500: Asum; % maximum raffle ticket number var 2..999: max_num; constraint A2 == A1 + Adiff; constraint A3 == A2 + Adiff; constraint A4 == A3 + Adiff; constraint A5 == A4 + Adiff; constraint Asum = A1 + A2 + A3 + A4 + A5; constraint G2 == G1*Gratio; constraint G3 == G2*Gratio; constraint G4 == G3*Gratio; constraint G5 == G4*Gratio; constraint Gsum = G1 + G2 + G3 + G4 + G5; constraint Gsum == Asum; constraint all_different ([A1,A2,A3,A4,A5,G1,G2,G3,G4,G5]); constraint max_num = max({A1,A2,A3,A4,A5,G1,G2,G3,G4,G5}); solve satisfy; output[ "Geometric Series: " ++ show([G1,G2,G3,G4,G5]) ++ "\n" ++ "Arithmetic Series = " ++ show([A1,A2,A3,A4,A5]) ++ "\n" ++ "Max ticket number = " ++ show(max_num) ];This programme produces three solutions, all with the same maximum value.
In only one of the solutions do the digits add to a perfect square.
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Jim Randell 8:05 pm on 11 December 2020 Permalink |
@Geoff: There’s only one copy of each ticket, so both George and Martha’s sets of numbers are fully determined. (Even though we are only asked for the highest numbered ticket).
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Frits 2:30 pm on 12 December 2020 Permalink |
The result of this piece of code are all numbers ending on a 1. This limits George’s lowest-numbered ticket to one number.
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Frits 7:57 pm on 12 December 2020 Permalink |
The generated code can be seen with option verbose=256.
from enigma import SubstitutedExpression, is_prime, is_square, \ seq_all_different # Formula # G * (1 + K + K^2 + K^3 + K^4) == 5 * (M + 2*D) # (1 + K + K^2 + K^3 + K^4) equals (K^5 - 1) / (K - 1) # (idea Brian Gladman) # sum the digits of the numbers in a sequence dsums = lambda seq: sum(sum(int(x) for x in str(s)) for s in seq) # domain lists dlist = list(range(3)) # d < 250 elist = list(range(10)) flist = list(range(10)) glist = list(range(1, 10)) klist = [] mlist = list(range(1, 10)) lastdigit = set() # last digit of the total of 5 tickets # K^4 may not be higher than 1000, so K < 6 for k in range(2, 6): t = sum(k**i for i in range(5)) klist.append(k) lastdigit.add(t % 10) lastdigit = list(lastdigit) if 5 not in lastdigit: # George's lowest ticket must be 5 as total is a multiple of 5 glist = [5] if len(lastdigit) == 1: # Martha's tickets sum to 5 * (M + 2*D) if lastdigit[0] % 2 == 1: # Martha's lowest ticket must be odd mlist = [1, 3, 5, 7, 9] else: # Martha's lowest ticket must be even mlist =[2, 4, 6, 8] if len(glist) == 1: # George's highest ticket may not be higher than 1000 klist = [x for i, x in enumerate(klist, start=2) if i**4 * glist[0] <= 1000] # calculate highest sum of 5 tickets t = sum(max(klist)**i for i in range(5)) * glist[0] # Martha's highest ticket may not be higher than (t/5 - M) / 2 dlist = [x for x in dlist if x <= (t / 10) // 100] if dlist == [0]: elist = [x for x in elist if x <= (t // 10) // 10] mlist = [x for x in mlist if x != glist[0]] # build dictionary for invalid digits vars = "DEFGKM" invalid = "dict(" for i in range(10): txt = "" for j, li in enumerate([dlist, elist, flist, glist, klist, mlist]): if i not in li: txt += vars[j] invalid += "[(" + str(i) + ", '" + txt + "')] + " invalid = invalid[:-3] + ")" exprs = [] # George's and Martha's sum was the same exprs.append("G * (K ** 5 - 1) // (K - 1) == 5 * (M + 2 * DEF)") # all ticket numbers have to be different exprs.append("seq_all_different([G, G * K, G * K**2, G * K**3, G * K**4, \ M, M + DEF, M + 2*DEF, M + 3*DEF, M + 4*DEF])") # total of all the digits in the ten numbers was a perfect square exprs.append("is_square(dsums([G, G * K, G * K**2, G * K**3, G * K**4, \ M, M + DEF, M + 2*DEF, M + 3*DEF, M + 4*DEF]))") # the alphametic puzzle p = SubstitutedExpression( exprs, answer="(G, G * K, G * K**2, G * K**3, G * K**4), \ (M, M + DEF, M + 2*DEF, M + 3*DEF, M + 4*DEF), 5 * (M + 2 * DEF)", d2i=eval(invalid), env=dict(dsums=dsums), distinct="GM", verbose=0) # Print answers for (_, ans) in p.solve(): print(f"{ans}")LikeLike
Tony Brooke-Taylor 9:10 am on 14 December 2020 Permalink |
My first solution generated the series explicitly and then applied the tests to each series, but then I realised we could equate the expressions for the sums of each series to reduce the sets of parameters we need to test. My second attempt was similar to below but instead of constraining the sum to a multiple of 5 I initially constrained my ‘b’ parameter to an integer, which I think is mathematically equivalent but came inside another loop so was less efficient.
#Generator function to create possible combinations of parameters for George and Martha def core_parameters(): #George's tickets form the geometric progression x.(y^0, y^1, y^2, y^3, y^4) #Martha's tickets form the arithmetic progression (a+b*0, a+b*1, a+b*2, a+b*3, a+b*4) for x in range(1,10): #we are told that George's first ticket has a single-digit value max_y = int((1000/x)**(0.25)//1)+1 #defined by the maximum possible value for George's highest-value ticket for y in range(2,max_y): #y=1 not possible because this would give all George's tickets the same value sum_values = x*(1-y**5)/(1-y) #sum of finite geometric progression #sum_values = a*5 + b*10 #sum of finite arithmetic progression #=> b = (sum_values - a*5)/10 = sum_values/10 - a/2 if sum_values%5 == 0: #equality also requires that the sum is a multiple of 5 for a in range(1,10): #we are told that Martha's first ticket has a single-digit value if a != x: #Martha's first ticket cannot have the same value as George's b = sum_values/10 - a/2 yield x, y, a, b #Function to sum all digits in an integer def digit_sum(num): s = 0 for dig in str(num): s = s + int(dig) return s #Control program for first_george_ticket, george_multiple, first_martha_ticket, martha_multiple in core_parameters(): george_tix = [first_george_ticket*george_multiple**n for n in range(5)] martha_tix = [int(first_martha_ticket+martha_multiple*m) for m in range(5)] #they can't both have the same ticket and we are told that the sum of all digits is a square if set(martha_tix).intersection(george_tix) == set() and \ ((sum([digit_sum(j) for j in george_tix]) + sum([digit_sum(k) for k in martha_tix]))**(1/2))%1 == 0: print("George:", george_tix) print("Martha:", martha_tix) print("Highest-valued ticket:",max(george_tix + martha_tix)) breakLikeLike