Sunday Times Teaser 3321 – The More you Buy, the More you Save
by Colin Vout
Published Sunday May 17 2026 (link)
… or so we are often encouraged to believe.
I needed to buy exactly 42 coloured pens and I could buy them in packs of 1, 4, 7, 11 or 17. A pack of 1 cost £2, a pack of 17 cost £23, and the prices of the other packs were whole numbers of pounds. The average price per pen always decreased going to larger packs. Given the prices I was faced with, I found a suitable and unique combination of the packs that gave me the lowest possible total cost, and this involved buying at least one pack of 11.
In ascending order of pack size, how many of each pack did I buy, and what were the pack prices (eg, 5 at £2 each, 3 at £8 each, etc)?
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BRG permalink1234567891011121314151617181920212223242526272829303132333435363738394041424344454647from itertools import productfrom collections import defaultdict# pens per packpacks = (1, 4, 7, 11, 17)# split <pens> pens into packsdef split(pens, pck=0, seq=()):if pck == len(packs):if not pens:yield seqelse:for np in range(pens // packs[pck] + 1):if (rem := pens - np * packs[pck]) >= 0:yield from split(rem, pck + 1, seq + (np,))# store splits for 42 penssplits = list(split(42))# consider the cost of packs of 4, 7 and 11 pensfor c3 in product(range(2, 23), repeat=3):cpp = (2,) + c3 + (23,)# find the average cost of pens for each packavp = [c / n for c, n in zip(cpp, packs)]# ensure this decreases with increasing pack sizesif all(a > b for a, b in zip(avp, avp[1:])):# store the minimum costs and the related solutionsmin_cost, sols = 1000, defaultdict(list)for ps in splits:cost = sum(p * c for p, c in zip(ps, cpp))if cost <= min_cost:if cost < min_cost:min_cost = costsols = defaultdict(list)sols[cost].append((ps, cpp))# look for the minimum unique cost with at least one 11 packfor cost, ss in list(sols.items()):if len(ss) == 1 and ss[0][1][3]:ps, cpp = ss[0]if ps[3]:fs = ', '.join(f"{p} at £{c}" for p, c in zip(ps, cpp) if p)f, s, l = fs.rpartition(', ')print(f"{f} and {l} (cost = £{cost}).")
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Frits permalink12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849from itertools import productpacks = [1, 4, 7, 11, 17]idx = {packs[i]: i for i in range(len(packs))}# decompose <t> into numbers from <ns> so that sum(chosen numbers) equals <t>def decompose(t, ns, s=[]):if t == 0:yield selse:for n in ns:if n > t: breakif not s or n <= s[-1]:yield from decompose(t - n, ns, s + [n])# determine possible quantities that add to 42qss = list(decompose(42, packs))# determine price ranges for packs 4, 7 and 11 (prices between 23/17 and 2)rngs = [[m for m in range(3, 23) if 17 * m > 23 * packs[i] andm < 2 * packs[i]] for i in range(1, 4)]pss = []# choose prices for packs 4, 7 and 11for p4, p7, p11 in product(*rngs):# the average price per pen always decreased going to larger packsif p4 * packs[2] <= p7 * packs[1]: continueif p7 * packs[3] <= p11 * packs[2]: continueps = (2, p4, p7, p11, 23)mn, mns = 9999, []# for all quantities that sum to 42for qs in qss:# check total costt = sum(ps[idx[q]] for q in qs)if t <= mn:if t < mn:mns = []mns.append((t, qs))mn = t# check for unique solution buying at least one pack of 11if len(mns) == 1 and 11 in (qs := mns[0][1]):# determine # at # eachq_at_p = [str(qs.count(q)) + ' at £' + str(ps[idx[q]]) + ' each'for q in sorted(set(qs))]print(f"answer: {', '.join(q_at_p)}")
