Sunday Times Teaser 3156 – Balancing Act
by Peter Good
Published Sunday March 19 2023 (link)
A circus elephant standing on one end of a see-saw pivoted at the centre was balanced by a troupe of fewer than 20 acrobats, each of equal weight, standing on the other end. The elephant moved forwards and several acrobats jumped off to maintain balance. The elephant moved backwards and some of them climbed back on to the end to rebalance.
The elephant always moved a prime number of feet and there was always a prime number of acrobats on the see-saw. If I told you how far backwards the elephant moved you could work out the numbers of acrobats.
(In equilibrium, the product of weight and distance from pivot point must be the same on both sides)
How far did the elephant move backwards, and how many acrobats are there in the troupe?
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BRG permalink12345678910111213141516171819202122232425262728293031# Acrobat numbers: (p, q, r) all prime with q < r < p < 20# Elephant positions: (e, e + f, e + f – b) with f and b prime with b < f e.(p – q) == p.f## p.(e – f + b) == r.e ==> e.(p – r) == p.(f – b)## f / (p – q) == (f – b) / (p – r) ==> f.(r – q) == b.(p – q)from itertools import combinations# primes for acrobat numbers on the see-saw and Elephant movesprms = (2, 3, 5, 7, 11, 13, 17, 19)sols = {}# consider possible primes for acrobat numbers …for q, r, p in combinations(prms, 3):# … and the elephant’s backward movefor b in prms:# calculate the forward move and check its validityf, _r = divmod(b * (p – q), r – q)if not _r and f in prms:sols[b] = sols.get(b, []) + [(p, q, r, f)]# look for a backward move that identifies a unique solutionfor b, l_pqrf in sols.items():(p, q, r, f), *_r = l_pqrfe = p * f / (p – q)e = e if e != (z := int(e)) else zif not _r:print(f”{p} Acrobats; Elephant @({e}) ==> {f} @({q}) <== {b} @({r})")
