Sunday Times Teaser 3326 – Interesting Figures

by BRG on 21 June 2026

By Andrew Gibbons

Published Sunday June 21 2026

At a cricket match, Tom Leggie bowled fewer than 100 overs (O), of which fewer than 10 were maidens (M), took more than one but fewer than 11 wickets (W), and had an average of runs conceded per over (R/O) that was less than 6. I noticed something interesting about his bowling analysis (O-M-R-W): writing the numbers together with no gaps or overlap, then reading from right to left, they comprised a perfect square followed immediately by a prime containing the same number of digits, and they could also be read (again in their entirety, right to left) as a prime number followed by a perfect square number containing more digits. None of the perfect squares or primes was a single digit, and no zeros were involved.

What was Tom’s bowling analysis (O-M-R-W)?

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BRG permalink


# produce two and three digit primes with no zero digits
prms = {3, 5, 7}
prms |= {2} | {n for n in range(11, 100, 2) if all(n % p for p in prms)}
prms2d = {str(p) for p in prms if p >= 10 and p % 10}
prms3d = {sp for n in range(101, 1000, 2) if all(n % p for p in prms)
                  and '0' not in (sp:= str(n))}

# produce three and four digit squares with no zero digits
sqrs3d = {sq for s in range(10, 32) if '0' not in (sq := str(s * s))}
sqrs4d = {sq for s in range(32, 100) if '0' not in (sq := str(s * s))}

# consider reversed values as three digit squares followed by 3 digit primes
for s in sqrs3d:
  for p in prms3d:
    rev = s + p
    # now consider them as two digit primes followed by four digit squares
    if (pp := rev[:2]) in prms2d and (ss := rev[2:]) in sqrs4d:
      ans = rev[::-1]
      for i, j in ((1, 2), (2, 3)):
        O, M, R, W = int(ans[:i]), int(ans[i:j]), int(ans[j:5]), int(ans[5])
        if R < 6 * O and W > 1:
          print(f"OMRW = {ans} [(O, M, R, W) = ({O}, {M}, {R}, {W}) "
                f"({s}, {p}), ({pp}, {ss})]")

# produce two and three digit primes with no zero digits

prms = {3, 5, 7}

prms |= {2} | {n for n in range(11, 100, 2) if all(n % p for p in prms)}

prms2d = {str(p) for p in prms if p >= 10 and p % 10}

prms3d = {sp for n in range(101, 1000, 2) if all(n % p for p in prms)

and '0' not in (sp:= str(n))}

# produce three and four digit squares with no zero digits

sqrs3d = {sq for s in range(10, 32) if '0' not in (sq := str(s * s))}

sqrs4d = {sq for s in range(32, 100) if '0' not in (sq := str(s * s))}

# consider reversed values as three digit squares followed by 3 digit primes

for s in sqrs3d:

for p in prms3d:

rev = s + p

# now consider them as two digit primes followed by four digit squares

if (pp := rev[:2]) in prms2d and (ss := rev[2:]) in sqrs4d:

ans = rev[::-1]

for i, j in ((1, 2), (2, 3)):

O, M, R, W = int(ans[:i]), int(ans[i:j]), int(ans[j:5]), int(ans[5])

if R < 6 * O and W > 1:

print(f"OMRW = {ans} [(O, M, R, W) = ({O}, {M}, {R}, {W}) "

f"({s}, {p}), ({pp}, {ss})]")

John Z. permalink


from number_theory import is_prime

# consider number of maidens
for M in range(1, 10): # fewer than 10 were maidens
  # consider number of overs
  for O in range(M, 100): # fewer than 100 overs
    # consider number of wickets
    for W in range(2, 10):
      # consider number of runs
      for R in range(O * 6):

        # concatenate and invert
        OMRWi = (str(O) + str(M) + str(R) + str(W))[::-1]

        # no zeroes; length must be even and at least 6
        if '0' in OMRWi or (L := len(OMRWi)) & 1 or L < 6:
          continue

        #potential square, potential prime
        ps, pp = int(OMRWi[:L // 2]), int(OMRWi[L // 2:])

        if int(ps ** 0.5) ** 2 == ps and is_prime(pp):
          # consider length of second prime number
          for pl in range(2, L // 2):

            # second test: prime then square
            pp2, ps2 = int(OMRWi[:pl]), int(OMRWi[pl:])

            if is_prime(pp2) and int(ps2 ** 0.5) ** 2 == ps2:
              print(O, M, R, W)
              print(ps, pp)
              print(pp2, ps2)

from number_theory import is_prime

# consider number of maidens

for M in range(1, 10): # fewer than 10 were maidens

# consider number of overs

for O in range(M, 100): # fewer than 100 overs

# consider number of wickets

for W in range(2, 10):

# consider number of runs

for R in range(O * 6):

# concatenate and invert

OMRWi = (str(O) + str(M) + str(R) + str(W))[::-1]

# no zeroes; length must be even and at least 6

if '0' in OMRWi or (L := len(OMRWi)) & 1 or L < 6:

continue

#potential square, potential prime

ps, pp = int(OMRWi[:L // 2]), int(OMRWi[L // 2:])

if int(ps ** 0.5) ** 2 == ps and is_prime(pp):

# consider length of second prime number

for pl in range(2, L // 2):

# second test: prime then square

pp2, ps2 = int(OMRWi[:pl]), int(OMRWi[pl:])

if is_prime(pp2) and int(ps2 ** 0.5) ** 2 == ps2:

print(O, M, R, W)

print(ps, pp)

print(pp2, ps2)

John Z. permalink

When someone (i.e. Brian) produces a great solution it’s easy to come up with some tweaks which may be an improvement. Here’s a stab at that:


# Given the constraints of the Teaser,
# we can have OMRRW (5 digits) or OOMRRW (6 digits) or OOMRRRW (7 digits).
# As the number of digits must be even there must be 6 digits: OOMRRW.

# produce two and three digit primes with no zero digits
prms = {3, 5, 7}
prms |= {n for n in range(11, 100, 2) if all(n % p for p in prms)}
prms2d = {str(p) for p in prms if p >= 10}
prms3d = {sp for n in range(101, 1000, 2) if all(n % p for p in prms)
                  and '0' not in (sp:= str(n))}

# produce 3 and 4 digit squares containing primes as first 2 or last 3 digits
sqrs3d = {sq for s in range(10, 32) if (sq := str(s * s))[:-1] in prms2d}
sqrs4d = {sq for s in range(32, 100) if (sq := str(s * s))[1:] in prms3d}

# consider three digit squares (containing embedded 2 digit primes)
for s in sqrs3d:
  # consider four digit squares containing embedded three digit primes
  for s4p3 in sqrs4d:
  
    # reversed values are three digit squares followed by 3 digit primes
    rev = s + s4p3[1:] # s4p3[1:] = embedded three digit prime
    
    # 2 digit prime followed by potential 4 digit square
    pp, ss = rev[:2], rev[2:]
    
    # are the last four digits a square? (we know the first 2 digits == prime)
    if ss in sqrs4d:
      ans = rev[::-1]

      # based on 6 digits in O-M-R-W i.e. OOMRRW
      O, M, R, W = int(ans[:2]), int(ans[2:3]), int(ans[3:5]), int(ans[5])
      if R < 6 * O and W >= M:
        print(f"OMRW = {ans} [(O, M, R, W) = ({O}, {M}, {R}, {W}) "
              f"({s}, {s4p3[1:]}), ({pp}, {ss})]")

# Given the constraints of the Teaser,

# we can have OMRRW (5 digits) or OOMRRW (6 digits) or OOMRRRW (7 digits).

# As the number of digits must be even there must be 6 digits: OOMRRW.

# produce two and three digit primes with no zero digits

prms = {3, 5, 7}

prms |= {n for n in range(11, 100, 2) if all(n % p for p in prms)}

prms2d = {str(p) for p in prms if p >= 10}

prms3d = {sp for n in range(101, 1000, 2) if all(n % p for p in prms)

and '0' not in (sp:= str(n))}

# produce 3 and 4 digit squares containing primes as first 2 or last 3 digits

sqrs3d = {sq for s in range(10, 32) if (sq := str(s * s))[:-1] in prms2d}

sqrs4d = {sq for s in range(32, 100) if (sq := str(s * s))[1:] in prms3d}

# consider three digit squares (containing embedded 2 digit primes)

for s in sqrs3d:

# consider four digit squares containing embedded three digit primes

for s4p3 in sqrs4d:

# reversed values are three digit squares followed by 3 digit primes

rev = s + s4p3[1:] # s4p3[1:] = embedded three digit prime

# 2 digit prime followed by potential 4 digit square

pp, ss = rev[:2], rev[2:]

# are the last four digits a square? (we know the first 2 digits == prime)

if ss in sqrs4d:

ans = rev[::-1]

# based on 6 digits in O-M-R-W i.e. OOMRRW

O, M, R, W = int(ans[:2]), int(ans[2:3]), int(ans[3:5]), int(ans[5])

if R < 6 * O and W >= M:

print(f"OMRW = {ans} [(O, M, R, W) = ({O}, {M}, {R}, {W}) "

f"({s}, {s4p3[1:]}), ({pp}, {ss})]")

BRG permalink

Hi John,

I did conclude that the rarity of squares when compared with primes would make for a faster solution but didn’t go further with this so it good to see that you have followed this approach up.

I also feel my original code is somewhat ‘over engineered’ and deserves simplification so I have come up with this as an alternative:


# produce two and three digit primes
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)}

# four digit squares without zero digits indexed on their most significant
# digit and for which their least significant three digits are primes  
s2sppp = dict((ss // 1000, ss) for s in range(32, 100)
              if (ss := s * s) % 1000 in prms and '0' not in str(ss))

# consider three digit squares ...
for s in range(10, 32):
  # ... with no zero digits and their top two digits forming primes
  if (ss := s * s) // 10 in prms and '0' not in str(ss):
    
    # use the common digit in the two representations to splice 
    # them to give the reversed six digit value and the answer
    rev = 10000 * (ss // 10) + s2sppp[ss % 10]
    ans = str(rev)[::-1]
    
    # extract the O, M, R, W values and check them
    for i, j in ((1, 2), (2, 3)):
      O, M, R, W = int(ans[:i]), int(ans[i:j]), int(ans[j:5]), int(ans[5])
      if R < 6 * O and W > 1:
        sr = str(rev)
        print(f"OMRW = {ans} [(O, M, R, W) = ({O}, {M}, {R}, {W}) "
              f"({sr[:3]}, {sr[3:]}), ({sr[:2]}, {sr[2:]})]")

# produce two and three digit primes

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)}

# four digit squares without zero digits indexed on their most significant

# digit and for which their least significant three digits are primes

s2sppp = dict((ss // 1000, ss) for s in range(32, 100)

if (ss := s * s) % 1000 in prms and '0' not in str(ss))

# consider three digit squares ...

for s in range(10, 32):

# ... with no zero digits and their top two digits forming primes

if (ss := s * s) // 10 in prms and '0' not in str(ss):

# use the common digit in the two representations to splice

# them to give the reversed six digit value and the answer

rev = 10000 * (ss // 10) + s2sppp[ss % 10]

ans = str(rev)[::-1]

# extract the O, M, R, W values and check them

for i, j in ((1, 2), (2, 3)):

O, M, R, W = int(ans[:i]), int(ans[i:j]), int(ans[j:5]), int(ans[5])

if R < 6 * O and W > 1:

sr = str(rev)

print(f"OMRW = {ans} [(O, M, R, W) = ({O}, {M}, {R}, {W}) "

f"({sr[:3]}, {sr[3:]}), ({sr[:2]}, {sr[2:]})]")

Sunday Times Teaser 3326 – Interesting Figures

By Andrew Gibbons

Published Sunday June 21 2026

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