From The Sunday Times, 17th February 1980 [link]
Last summer, for a pleasant holiday pastime with mathematical connections, I took up the job of operating our local cricket scoreboard at matches.
This scoreboard is the envy of all the other teams in the league. It shows:
– the two batsmen identified by their numbers in the batting order, with their individual totals;
– the score (i.e. the number of runs and the wickets fallen);
– the number of overs bowled;
– the number of extras;
– the score of the last man out.
During one match, while the two batsmen were in full flight, our team declared their innings closed at the end of an over, with wickets to spare. Exhausted after a busy day, I examined the board and was amazed to see that all of the numbers recorded on it used only two different digits between them.
I noted that the total was the only three-figure number; that there were four two-figure numbers; and that two single-figure numbers appeared twice. I also observed that the score of the last man out was a factor of the total, and the division of the latter by the former still resulted in a single figure number on the board.
I was pleased to see that all the batsmen on the board reached double figures.
What was the final score, i.e. how many runs for how many wickets?
How many extras were there?
[In cricket, the batsmen are numbered from 1 upwards as they come in to bat. The world record is 77 runs in one over. The match itself was perfectly normal — no-one retired hurt etc. Ed.]
This puzzle is included in the book The Sunday Times Book of Brainteasers (1994). The puzzle text above is taken from the book.
[teaser917]
Frits and 
Hugo, 
NigelR, and 1 other are discussing. 
S2T2 
Jim Randell 10:12 am on 12 September 2023 Permalink |
If the area of the original field is A, then it is split such that the northern field has an area N = (2/7)A and the southern field S = (5/7)A.
The boundaries of N are all integers and so form a Pythagorean triple (x, y, z).
And we can expand the northern field by a factor of f until its hypotenuse exactly covers the boundary of the original field, and we get this:
The we can then write the area of the original field in two ways:
And we can calculate the unknown sides of S:
This Python program runs in 58ms. (Internal runtime is 334µs).
Run: [ @replit ]
from enigma import (pythagorean_triples, div, ihypot, ordered, printf) for (a, b, z) in pythagorean_triples(100): z_ = div(5 * z, 2) if z_ is None: continue N = ordered(a, b, z) for (x, y) in [(a, b), (b, a)]: y_ = div(ihypot(7 * x, 5 * y), 2) if y_ is None: continue S = ordered(x, z_, y_) if 25 in S: # output solution printf("N = {N}; S = {S}")Solution: The other two sides of the southern field are 6 and 29 furlongs.
N is a (6, 8, 10) triangle, (area = 24 square furlongs ≈ 0.97 sq km).
S is a (6, 25, 29) triangle, (area = 60 square furlongs ≈ 2.43 sq km).
So the original field was a (8, 29, 35) triangle, (area = 84 square furlongs ≈ 3.40 sq km).
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Frits 5:44 pm on 12 September 2023 Permalink |
Avoiding imports.
is_square = lambda n: int(rt := n**.5) == rt # generate Pythagoren triples for a fixed hypothenuse def triples(h): sqs = {i * i for i in range(1, h)} return [(int(s1**.5), int(s2**.5)) for s1 in sqs if (s2 := h**2 - s1) >= s1 and s2 in sqs] side = 25 # can x be 25? # then f.x = 87.5, y' can never be a whole number (f = 3.5) # if other side (2.5y) is k.0 or k.5, so x != 25 # can y' be 25? for c, d in triples(side): # are these lenghts multiples of 2.5 and 3.5? x, y = -1, -1 if (c % 2.5 == 0 and d % 3.5 == 0): y, x = c // 2.5, d // 3.5 if ((c % 3.5 == 0 and d % 2.5 == 0)): x, y = c // 3.5, d // 2.5 if x == -1: continue # y' can be 25 if is_square(z2 := x**2 + y**2): z = int(z2**.5) if z % 2 == 0: print(f"answer {int(x)} and {int(2.5 * z)} furlongs") # can z' be 25? if side % 5: exit(0) z = int(side * 2 / 5) ts = triples(z) # determine z' for x, y in ts + [x[::-1] for x in ts]: if is_square(y1 := 3.5**2 * x**2 + 2.5**2 * y**2): print(f"answer {int(x)} and {int(y1**.5)} furlongs")LikeLike