Sunday Times Teaser 3327 – Water, Water Everywhere

by BRG on 28 June 2026

by Danny Roth

Published Sunday June 28 2026

Having moved to the southern coast of a large lake, George and Martha have acquired a speedboat, capable of doing V kph. They are going to visit their daughter who lives 60km due north on the opposite shore of the lake. A steady current is flowing from the West due East. George has told Martha that he intends to propel the boat at top speed at such an angle that it will complete the direct south-north journey in T minutes. “In that case,” replied Martha confidently, “you have allowed for the current to be flowing at C kph, reducing your speed by exactly 1 or 2 kph to S.” C, V, S and T are all two-digit whole numbers, with S+T being a perfect power.

How fast is that current flowing?

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BRG permalink


# the speed of the speedboat
for V in range(10, 100):
  # the speed due north:
  for S in (V - 1, V - 2):
    if S < 10:
      continue
    # calculate the speed of the current (two digits)
    c2 = V * V - S * S
    if (C := round(c2 ** 0.5)) ** 2 == c2 and 10 <= C < 100:
      # calculate the two digit journey time
      T, r = divmod(3600, S) 
      if r or not 10 <= T <= 100:
        continue
      # look for an S + T value that is a perfect power
      p_pwr = S + T
      for p in range(2, 8):
        if (q := round(p_pwr ** (1 / p))) ** p == p_pwr:
          print(f"{C = } ({V = }, {S = }, {T = }, S + T = {q}^{p}).")

# the speed of the speedboat

for V in range(10, 100):

# the speed due north:

for S in (V - 1, V - 2):

if S < 10:

continue

# calculate the speed of the current (two digits)

c2 = V * V - S * S

if (C := round(c2 ** 0.5)) ** 2 == c2 and 10 <= C < 100:

# calculate the two digit journey time

T, r = divmod(3600, S)

if r or not 10 <= T <= 100:

continue

# look for an S + T value that is a perfect power

p_pwr = S + T

for p in range(2, 8):

if (q := round(p_pwr ** (1 / p))) ** p == p_pwr:

print(f"{C = } ({V = }, {S = }, {T = }, S + T = {q}^{p}).")

Frits permalink

One main loop, only six powers have to be considered.


# express S in terms of power P and we can deduce P >= 120

# perfect power n = a^b between 120 and 197 (99 + 98)
ps = {p for a in range(2, 15) for b in range(2, 8) 
      if 120 <= (p := a**b) <= 197}

# C^2 = 2.S + 1 or C^2 = 4.(S + 1) and S < 99
for C in range(10, int((4 * 99)**.5) + 1):
  S = (C**2 - 1) // 2 if C % 2 else (C**2 - 4) // 4
  # S > 36 otherwise T > 99
  if not (37 <= S < 100): continue
  T, r = divmod(3600, S)
  if r or (S + T) not in ps: continue
  print(f"answer: {C} km/h")

# express S in terms of power P and we can deduce P >= 120

# perfect power n = a^b between 120 and 197 (99 + 98)

ps = {p for a in range(2, 15) for b in range(2, 8)

if 120 <= (p := a**b) <= 197}

# C^2 = 2.S + 1 or C^2 = 4.(S + 1) and S < 99

for C in range(10, int((4 * 99)**.5) + 1):

S = (C**2 - 1) // 2 if C % 2 else (C**2 - 4) // 4

# S > 36 otherwise T > 99

if not (37 <= S < 100): continue

T, r = divmod(3600, S)

if r or (S + T) not in ps: continue

print(f"answer: {C} km/h")

Sunday Times Teaser 3327 – Water, Water Everywhere

by Danny Roth

Published Sunday June 28 2026

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