Sunday Times Teaser 3284 – Cricket Square

by BRG on August 31, 2025

by Andrew Skidmore

Published Sunday August 31 2025 (link)

We recently played a thrilling cricket match against the league leaders. Each team had two innings alternately (our opponents going first) and the combined total of runs scored in each innings determined the winner. Curiously both our opponents’ scores were 3-figure squares, the six digits all being different. The total of those two scores was also a square.

After our first innings (in which we scored a 3-figure prime number of runs), we had a lead, but after their second innings we trailed by a 3-figure prime number (less than 250). The six digits in these two primes were all different. Our second innings score was also a 3-figure prime number but we lost the match by a prime number of runs.

How many runs did we score in our second innings?

From → Uncategorized

4 Comments Leave one →

BRG permalink


# squares in the range 100..2000
sqs = tuple(n ** 2 for n in range(10, 44))
# primes below 1000
prms = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
prms |= {2} | {n for n in range(37, 1000, 2) if all(n % p for p in prms)}
# primes in 100 .. 1000
prm3d = sorted((n for n in prms if n > 100))

# consider the possible scores in the opponent's first and second innings
for sq1 in sqs[:sqs.index(1024)]:
  for sq2 in sqs[:sqs.index(1024)]:
    # the opponents final score is a square and their two 
    # scores combined have six different digits
    if sq1 + sq2 not in sqs or len(set(str(sq1) + str(sq2))) != 6:
      continue
    
    # consider the possible scores in our first innings
    for pr1 in prm3d:
      # we had a first innings lead but we were behind afer 
      # their second innings by a prime less than 250 ...
      if pr1 > sq1 and (p := sq1 + sq2 - pr1) < 250 and p in prms:
        # ... and the two primes have six different digits
        if len(set(str(pr1) + str(p))) == 6:
        
          # consider the possible scores in our second innings
          for pr2 in prm3d:
            # we lost the match by a prime number of runs
            if sq1 + sq2 - (pr1 + pr2) in prms:
              print(f"{sq1} {pr1} {sq2} {pr2} -> {pr2}")

# squares in the range 100..2000

sqs = tuple(n ** 2 for n in range(10, 44))

# primes below 1000

prms = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31}

prms |= {2} | {n for n in range(37, 1000, 2) if all(n % p for p in prms)}

# primes in 100 .. 1000

prm3d = sorted((n for n in prms if n > 100))

# consider the possible scores in the opponent's first and second innings

for sq1 in sqs[:sqs.index(1024)]:

for sq2 in sqs[:sqs.index(1024)]:

# the opponents final score is a square and their two

# scores combined have six different digits

if sq1 + sq2 not in sqs or len(set(str(sq1) + str(sq2))) != 6:

continue

# consider the possible scores in our first innings

for pr1 in prm3d:

# we had a first innings lead but we were behind afer

# their second innings by a prime less than 250 ...

if pr1 > sq1 and (p := sq1 + sq2 - pr1) < 250 and p in prms:

# ... and the two primes have six different digits

if len(set(str(pr1) + str(p))) == 6:

# consider the possible scores in our second innings

for pr2 in prm3d:

# we lost the match by a prime number of runs

if sq1 + sq2 - (pr1 + pr2) in prms:

print(f"{sq1} {pr1} {sq2} {pr2} -> {pr2}")

Frits permalink


is_square = lambda n: round(n**(0.5))**2  ==  n

# 3-digit primes
P = [3, 5, 7]
P += [x for x in range(11, 32, 2) if all(x % p for p in P)]
P = [x for x in range(999, 100, -2) if all(x % p for p in P)]

sols = set()
# opponents first innings
for ab in range(10, 32): 
  oruns1 = ab * ab
  if len(set(s1 := str(oruns1))) != 3: continue
  # opponents second innings (total for opponents must be even)
  for de in range(10 + (ab % 2), 32, 2): 
    oruns2 = de * de
    # the total of those two scores was also a square
    if (t := oruns1 + oruns2) >= 1249: break
    if not is_square(t): continue
    # the six digits all being different
    if len(set(s1 + str(oruns2))) != 6: continue
    
    # setters first innings
    for sruns1 in P:
      # after the opponents second innings the setters team trailed
      # by a 3-digit prime number of runs (less than 250)
      if (trail :=  oruns1 + oruns2 - sruns1) >= 250: break
      if trail < 100 or trail not in P: continue
      # setters team took the lead after their first innings
      if not (sruns1 > oruns1): continue
      
      # the six digits all being different
      if len(set(str(sruns1) + str(trail))) != 6: continue
      
      # the setters team lost the match by a prime number of runs
      if (sruns2 := oruns1 + oruns2 - sruns1 - 2) not in P: continue
      sols.add(str(sruns2))

print(f"answer: {' or '.join(sols)} runs")

is_square = lambda n: round(n**(0.5))**2 == n

# 3-digit primes

P = [3, 5, 7]

P += [x for x in range(11, 32, 2) if all(x % p for p in P)]

P = [x for x in range(999, 100, -2) if all(x % p for p in P)]

sols = set()

# opponents first innings

for ab in range(10, 32):

oruns1 = ab * ab

if len(set(s1 := str(oruns1))) != 3: continue

# opponents second innings (total for opponents must be even)

for de in range(10 + (ab % 2), 32, 2):

oruns2 = de * de

# the total of those two scores was also a square

if (t := oruns1 + oruns2) >= 1249: break

if not is_square(t): continue

# the six digits all being different

if len(set(s1 + str(oruns2))) != 6: continue

# setters first innings

for sruns1 in P:

# after the opponents second innings the setters team trailed

# by a 3-digit prime number of runs (less than 250)

if (trail := oruns1 + oruns2 - sruns1) >= 250: break

if trail < 100 or trail not in P: continue

# setters team took the lead after their first innings

if not (sruns1 > oruns1): continue

# the six digits all being different

if len(set(str(sruns1) + str(trail))) != 6: continue

# the setters team lost the match by a prime number of runs

if (sruns2 := oruns1 + oruns2 - sruns1 - 2) not in P: continue

sols.add(str(sruns2))

print(f"answer: {' or '.join(sols)} runs")

BRG permalink

This version exploits the fact that two of the primes involved are a ‘prime pair’.


# Let sq1 and sq2 be the opponents square scores and pr1 and pr2 be our
# prime scores; hence with two further primes pr3 < 250 and pr4 we have:
#
#  s1 + s2 == pr1 + pr3 == pr1 + pr2 + pr4 ==> pr4 = pr3 - pr2
#
# Hence s1 + s2 is even, pr4 == 2 and (pr2, pr3) is a 'prime pair'

# squares in the range 100..2000
sqs = tuple(n ** 2 for n in range(10, 44))
# three digit primes
prms = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
prms = tuple(n for n in range(101, 1000, 2) if all(n % pr3 for pr3 in prms))
# prime pairs less than 250
pprs = tuple((p, p + 2) for p in prms if p + 2 in prms and p < 248)

# consider the possible scores in the opponent's first and second innings
for sq1 in sqs[:sqs.index(1024)]:
  for sq2 in sqs[:sqs.index(1024)]:
    # the opponents final score is an even square and their two scores combined
    # have six different digits 
    if (t := sq1 + sq2) % 2 or t not in sqs or len(set(str(sq1) + str(sq2))) != 6:
      continue
    
    # consider the prime pairs pr2 and pr3 and test that pr1 is prime
    for pr2, pr3 in pprs:
      # pr1 is prime and pr1 and pr3 have six different digits
      if ((pr1 := sq1 + sq2 - pr3) in prms and pr1 > sq1
                            and len(set(str(pr1) + str(pr3))) == 6):
        print(f"{sq1} {pr1} {sq2} {pr2} -> {pr2}")

# Let sq1 and sq2 be the opponents square scores and pr1 and pr2 be our

# prime scores; hence with two further primes pr3 < 250 and pr4 we have:

# s1 + s2 == pr1 + pr3 == pr1 + pr2 + pr4 ==> pr4 = pr3 - pr2

# Hence s1 + s2 is even, pr4 == 2 and (pr2, pr3) is a 'prime pair'

# squares in the range 100..2000

sqs = tuple(n ** 2 for n in range(10, 44))

# three digit primes

prms = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31}

prms = tuple(n for n in range(101, 1000, 2) if all(n % pr3 for pr3 in prms))

# prime pairs less than 250

pprs = tuple((p, p + 2) for p in prms if p + 2 in prms and p < 248)

# consider the possible scores in the opponent's first and second innings

for sq1 in sqs[:sqs.index(1024)]:

for sq2 in sqs[:sqs.index(1024)]:

# the opponents final score is an even square and their two scores combined

# have six different digits

if (t := sq1 + sq2) % 2 or t not in sqs or len(set(str(sq1) + str(sq2))) != 6:

continue

# consider the prime pairs pr2 and pr3 and test that pr1 is prime

for pr2, pr3 in pprs:

# pr1 is prime and pr1 and pr3 have six different digits

if ((pr1 := sq1 + sq2 - pr3) in prms and pr1 > sq1

and len(set(str(pr1) + str(pr3))) == 6):

print(f"{sq1} {pr1} {sq2} {pr2} -> {pr2}")

Frits permalink


# get 2 numbers that when squared sum up to n (with minimum m)
# https://stackoverflow.com/questions/72093402/sum-of-two-squares-in-python
def sum_of_two_squares(n, m):
  i = m
  j = int(n ** (1/2))
 
  while i < j:
    x = i * i + j * j
    if x == n:
      yield i, j
      i += 1
      j -= 1
    elif x < n:
      i += 1
    else:
      j -= 1
 
# 3-digit primes with different digits
P = {3, 5, 7}
P |= {x for x in range(11, 32, 2) if all(x % p for p in P)}
P = [x for x in range(103, 988, 2) if all(x % p for p in P)] 
sols = set()
# setters second innings
for srun2 in P:
  if (trail := srun2 + 2) >= 250: break
  if not (trail in P): continue
 
  # even squares for opponents total, (135 + 246) <= square < sqrt(987 + 249)
  for t in [i * i for i in range(20, 36, 2)]:
    # setters first innings
    if not ((srun1 := t - trail) in P): continue
 
    # the six digits all being different
    if len(set(str(srun1) + str(trail))) != 6: continue
 
    # the setters team lost the match by a prime number of runs
    if (srun2 := t - srun1 - 2) not in P: continue
 
    # get scores for opponents
    for a, b in sum_of_two_squares(t, 10):
      # the six digits all being different
      if len(set(str(sq1 := a * a) + str(sq2 := b * b))) != 6: continue
 
      # setters team took the lead after their first innings
      if srun1 <= sq1: continue
      sols.add(str(srun2))
        
print(f"answer: {' or '.join(sols)} runs")

# get 2 numbers that when squared sum up to n (with minimum m)

# https://stackoverflow.com/questions/72093402/sum-of-two-squares-in-python

def sum_of_two_squares(n, m):

i = m

j = int(n ** (1/2))

while i < j:

x = i * i + j * j

if x == n:

yield i, j

i += 1

j -= 1

elif x < n:

i += 1

else:

j -= 1

# 3-digit primes with different digits

P = {3, 5, 7}

P |= {x for x in range(11, 32, 2) if all(x % p for p in P)}

P = [x for x in range(103, 988, 2) if all(x % p for p in P)]

sols = set()

# setters second innings

for srun2 in P:

if (trail := srun2 + 2) >= 250: break

if not (trail in P): continue

# even squares for opponents total, (135 + 246) <= square < sqrt(987 + 249)

for t in [i * i for i in range(20, 36, 2)]:

# setters first innings

if not ((srun1 := t - trail) in P): continue

# the six digits all being different

if len(set(str(srun1) + str(trail))) != 6: continue

# the setters team lost the match by a prime number of runs

if (srun2 := t - srun1 - 2) not in P: continue

# get scores for opponents

for a, b in sum_of_two_squares(t, 10):

# the six digits all being different

if len(set(str(sq1 := a * a) + str(sq2 := b * b))) != 6: continue

# setters team took the lead after their first innings

if srun1 <= sq1: continue

sols.add(str(srun2))

print(f"answer: {' or '.join(sols)} runs")

Sunday Times Teaser 3284 – Cricket Square

by Andrew Skidmore

Published Sunday August 31 2025 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta