Sunday Times Teaser 3303 – Knights and Knaves
by John Owen
Published Sunday January 11 2026 (link)
On my travels, I came across a group of some knights (always truthful) and some knaves (always lying). At first, I couldn’t tell which were knights and which were knaves. However, six individuals then gave the following statements about their group: “The number of knights is prime.” “The number of knights is odd.” “The number of knaves is square.” “The number of knaves is even.” “The total of knights and knaves is an odd number.” “There are more knaves than knights.”
These statements enabled me to deduce the composition of the group.
How many (a) knights and (b) knaves were in the group?
2 Comments
Leave one →
-
BRG permalink12345678910111213141516171819202122232425262728293031# the combined number of knights and knavesgroup_size = 2while True:group_size += 1sols = set()for knights in range(1, group_size):knaves = group_size - knightsans = [# The number of knights is primeknights in {2, 3, 5, 7, 11, 13},# The number of knights is oddbool(knights & 1),# The number of knaves is squareround(knaves ** 0.5) ** 2 == knaves,# The number of knaves is evenknaves % 2 == 0,# The total of knights and knaves is an odd numberbool((knights + knaves) & 1),# There are more knaves than knightsknaves > knights]# are answers consistent with the numbers of knights and knavesif ans.count(True) <= knights and ans.count(False) <= knaves:sols.add((knights, knaves))# can we deduce the group composition?if len(sols) == 1:knights, knaves = sols.pop()print(f"{knights} knights and {knaves} knaves.")break
-
Frits permalink123456789101112131415161718192021222324252627282930313233343536373839from number_theory import is_primeis_square = lambda n: round(n**(0.5))**2 == n# check answers for the number of knives and knavesdef check(kni, kna):# count number of True statementsn_true = sum([is_prime(kni),kni % 2,is_square(kna),kna % 2 == 0,(kni + kna) % 2,kna > kni])return (n_true, 6 - n_true)# total of knights and knavesn = 6while True:sols = []for n_knights in range(1,n):n_knaves = n - n_knights# detect when there are multiple solutions for sureif n > 12 and n_knights >= 6 and n_knaves >= 6: continue# check the six statements(n_t, n_f) = check(n_knights, n_knaves)# collect possible solutionsif n_t <= n_knights and n_f <= n_knaves:sols.append((n_knights, n_knaves))# these statements enabled me to deduce ....if len(sols) == 1:print("answer", sols[0])n += 1if n > 999: # weak spot: no analysis have been done to limit nbreak
