Teaser 3179: Sums and products
From The Sunday Times, 27th August 2023 [link] [link]
Mrs Green uses a dice game to improve the maths skills of her pupils. The children sit four to a table and each child throws five dice. Points are awarded according to the sum and product for each child; bonus points are awarded if the five scores contain three or more of the same number.
At Liam’s table all four children had the same sum but all achieved that sum in different ways. Surprisingly, all of the twenty dice scored more than one and no bonus points were awarded. The distribution of scores was:
sixes … 5;
fives … 4;
fours … 2;
threes … 4;
twos … 5.Liam had the highest product.
What, in ascending order, were Liam’s five scores?
[teaser3179]
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Jim Randell 4:54 pm on 25 August 2023 Permalink |
A bit more straightforward than the last couple of puzzles.
This Python program runs in 52ms. (Internal runtime is 457µs).
from enigma import (subsets, multiset, multiply, ediv, printf) # required distribution of values dist = multiset.from_dict({ 6: 5, 5: 4, 4: 2, 3: 4, 2: 5 }) # find the common sum <k> for the 4 children k = ediv(dist.sum(), 4) # generate possible scores def scores(dist, k): # no value occurs more than 2 times m = multiset.from_pairs((k, min(2, v)) for (k, v) in dist.items()) # look for scores from 5 dice, with the required sum for ss in subsets(sorted(m), size=5, select='uC'): if sum(ss) == k: yield ss # consider 4-sets of scores for ss in subsets(scores(dist, k), size=4): # look for required distribution if multiset.from_seq(*ss) == dist: printf("sum = {k} -> scores = {ss}") # find the set with the max product ans = max(ss, key=multiply) # output solution printf("ans = {ans}")Solution: Liam’s scores were: 3, 3, 4, 5, 5.
Each child on Liam’s table threw 5 dice with a sum of 20:
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Frits 9:25 pm on 25 August 2023 Permalink |
from enigma import SubstitutedExpression # (A, B, C, D, E), (F, G, H, I, J), (K, L, M, N, O), (P, Q, R ,S, T) # are four rows of frequencies (0-2) # the alphametic puzzle p = SubstitutedExpression( [ # check first 3 scores of 5 throws "5 - A - B - C - D = E", "2 * A + 3 * B + 4 * C + 5 * D + 6 * E == 20", "F >= A", "5 - F - G - H - I = J", "2 * F + 3 * G + 4 * H + 5 * I + 6 * J == 20", "K >= F", "5 - K - L - M - N = O", "2 * K + 3 * L + 4 * M + 5 * N + 6 * O == 20", # fill up to target frequency "5 - A - F - K = P", "P >= K", "4 - B - G - L = Q", "2 - C - H - M = R", "4 - D - I - N = S", "5 - E - J - O = T", "2 * P + 3 * Q + 4 * R + 5 * S + 6 * T == 20", # different throws, avoid duplicates "sorted(set(@rows)) == @rows", ], answer="[(multiply(i**y for i, y in enumerate(x, 2) if y), x) \ for x in @rows]", base=3, d2i="", macro=dict([("rows", "[(A, B, C, D, E), (F, G, H, I, J), \ (K, L, M, N, O), (P, Q, R ,S, T)]")]), distinct="", digits=range(0, 3), reorder=0, verbose=0, # use 256 to see the generated code ) # print answers for ans in p.answers(): nums = [[i] * y for i, y in enumerate(ans[0][1], 2) if y] print(f"answer: {sorted(y for x in nums for y in x)}")LikeLike