Teaser 2332: Schoolwork
From The Sunday Times, 3rd June 2007 [link]
Five girls were each given £4 to spend on rubbers, pencils and pens (not necessarily all three). Rubbers cost 20p each, pencils 30p, pens 40p. They each bought 12 items, spending the entire £4.
Sue bought the same number of pencils as Tina did rubbers. Tina had as many pens as Ursula had pencils. Viola had as many rubbers as Wendy had pens. And just one girl had the same number of pens as Tina had pencils.
How many of each item did Ursula buy?
[teaser2332]
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Jim Randell 4:45 pm on 4 September 2025 Permalink |
Here is a solution using the [[
SubstitutedExpression]] solver from the enigma.py library.It runs in 129ms. (Internal runtime of the generated program is 1.5ms).
Solution: Ursula bought: 1 rubber; 6 pencils; 5 pens.
The full solution is:
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Jim Randell 11:24 am on 5 September 2025 Permalink |
Here is a solution using the [[
choose()]] function from the enigma.py library.It runs in 70ms (with an internal runtime of 387µs).
from enigma import (Record, decompose, choose, icount_exactly as count, nl, printf) # look for combinations of 12 items with a cost of 400p vs = list() for (rb, pl, pn) in decompose(12, 3, increasing=0, sep=0, min_v=0): if 2 * rb + 3 * pl + 4 * pn == 40: vs.append(Record(rb=rb, pl=pl, pn=pn)) # selection functions (in order) for each girl fns = [ None, # S (lambda S, T: S.pl == T.rb), # T (lambda S, T, U: T.pn == U.pl), # U None, # V (lambda S, T, U, V, W: V.rb == W.pn), # W ] # find candidate solutions for (S, T, U, V, W) in choose(vs, fns): # "just one girl bought the same number of pens as T did pencils" if not count((S, T, U, V, W), (lambda x, t=T.pl: x.pn == t), 1): continue # output solution printf("S: {S}{s}T: {T}{s}U: {U}{s}V: {V}{s}W: {W}", s=nl) printf()LikeLike
Frits 6:36 pm on 4 September 2025 Permalink |
from itertools import product N, T = 12, 400 pa, pb, pc = 20, 30, 40 # each girl spent 400p and bought 12 items and # bought <a> rubbers, <b> pencils and <c> pens # quantity options for the girls sols = [(a, b, c) for a in range(min(N, T // pa) + 1) for b in range(min(N - a, (T - a * pa) // pb) + 1) if T - pa * a - pb * b == pc * (c := N - a - b)] # determine quantities for all girls for S, T, U, V, W in product(sols, repeat=5): # Sue bought the same number of pencils as Tina did rubbers if S[1] != T[0]: continue # Tina had as many pens as Ursula had pencils if T[2] != U[1]: continue # Viola had as many rubbers as Wendy had pens if V[0] != W[2]: continue # just one girl bought the same number of pens as T did pencils if [x[2] for x in (S, T, U, V, W)].count(T[1]) != 1: continue print("answer:", U)LikeLike
Ruud 7:43 am on 5 September 2025 Permalink |
import types combinations = [ types.SimpleNamespace(rubbers=rubbers, pencils=pencils, pens=pens) for rubbers in range(20) for pencils in range(14) for pens in range(11) if pens * 40 + pencils * 30 + rubbers * 20 == 400 and pens + pencils + rubbers == 12 ] for s in combinations: for t in combinations: if s.pencils == t.rubbers: for u in combinations: if t.pens == u.pencils: for v in combinations: for w in combinations: if v.rubbers == w.pens: if sum(girl.pens == t.pencils for girl in (s, t, u, v, w)) == 1: for name in "Sue Tina Ursula Viola Wendy".split(): print(f"{name:7} {locals()[name[0].lower()]}")LikeLike