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]
GeoffR and 
John Crabtree and 
S2T2 
Jim Randell 8:14 am on 20 February 2026 Permalink |
This Python program runs in 69ms. (Internal runtime is 1.6ms)
from enigma import (irange, decompose, tuples, csum, multiply, is_square_p, printf) # the sequence of digits digits = list(irange(1, 9)) n = len(digits) # divide the digits into k groups (min groups = 1, max groups = n) for k in irange(1, n): for js in decompose(len(digits), k, increasing=0, sep=0, min_v=1): # split the digits into the groups gs = list(digits[i:j] for (i, j) in tuples(csum(js, empty=1), 2)) # calculate the result of the process r = multiply(sum(ns) for ns in gs) # both B and J found a square number if is_square_p(r): # B's was odd if r % 2 == 1: printf("Billy: {gs} -> {r}") # J's groups each have more than one digit if all(len(ns) > 1 for ns in gs): printf("Jilly: {gs} -> {r}")Solution: Billy = 12-3-456-78-9; Jilly = 12-3456-789.
So:
Here is a complete list of square numbers that can be formed this way:
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ruudvanderham 12:07 pm on 20 February 2026 Permalink |
I make expressions by putting either )*( or + between the digits 1 – 8, like
(1)*(2+3+4+5)*(6+7)*(8)*(9)
and then evaluate that with eval.
For Jilly, I then also check whether the lengths of all parts split by ‘)*(‘ is greater than 2.
import itertools import math for ops in itertools.product(("+", ")*("), repeat=8): exp = "(1" + "".join(f"{op}{i}" for i, op in enumerate(ops, 2)) + ")" if math.isqrt(val := eval(exp)) ** 2 == val: if val % 2: print("Billy", exp, "=", val) else: if all(len(part) > 2 for part in exp.split(")*(")): print("Jilly", exp, "=", val)LikeLike
Ruud 3:41 pm on 20 February 2026 Permalink |
With a more elegant check for no groups of one digit:
import itertools import math for ops in itertools.product(("+", ")*("), repeat=8): exp = "(1" + "".join(f"{op}{i}" for i, op in enumerate(ops, 2)) + ")" if math.isqrt(val := eval(exp)) ** 2 == val: if val % 2: print("Billy", exp, "=", val) else: if all("+" in part for part in exp.split("*")): print("Jilly", exp, "=", val)LikeLike