Sunday Times Teaser 3258 – On Grid
by Howard Williams
Published Sunday March 02 2025 (link)
Using all the digits 1 to 9 in a 3 by 3 grid, I have devised a game. In this grid, if each of the three rows, three columns and two diagonals are read both forwards and backwards, there are 16 three-digit numbers. Points are awarded to each of these three-digit numbers with 1 point awarded if it’s a prime number, 3 points if it’s a perfect square, 6 points if a perfect cube, with the points being cumulative if more than one applies.
What is the highest number of points that can be awarded?
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BRG permalink12345678910111213141516171819202122232425262728293031323334353637383940414243from itertools import permutationspr = {3, 5, 7}pr = {2} | pr | {n for n in range(11, 100, 2) if all(n % p for p in pr)}# three digit primes, squares and cubesprms = set(n for n in range(101, 1000, 2) if all(n % p for p in pr))sqrs = set(n * n for n in range(10, 32))cubs = set(n ** 3 for n in range(5, 10))# three digit numbers from three digits forward and reverseF = lambda x, y, z: 100 * x + 10 * y + zR = lambda x, y, z: 100 * z + 10 * y + x# for scorespts, rs = 0, set()# consider all combinations of the digits 1..9for p in permutations(range(1, 10)):if not (p[0] < min(p[2], p[6], p[8]) and p[1] < p[3]):continue# form rows, columns and diagonalsr = tuple((p[:3], p[3:6], p[6:]))c = tuple(zip(*r))d = (p[0:9:4], p[2:8:2])# create the sixteen three digit valuess16 = {*tuple(F(*s) for s in r + c + d),*tuple(R(*s) for s in r + c + d)}# find the scoret = sum(m * len(c & s16) for m, c in ((1, prms), (3, sqrs), (6, cubs)))# save the highest score and the associated grid valuesif t > pts:pts = trs = set()if t == pts:rs.add(r)# output the highest score and the associated gridsprint(f"The highest score is {pts} (grid rows: ", end='')for r in rs:print(f"{', '.join(''.join(str(d) for d in t) for t in r)}).")
