Sunday Times Teaser 2699 – Low Power

by Peter G Chamberlain

In the addition:

K I L O
+ W A T T
=======
H O U R

different letters consistently represent different digits and WATT represents a number that is a perfect power.

What three digit number does LOW represent?

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brian gladman permalink


from itertools import count, permutations

# return integer powers that match the pattern WATT
# (note: T is non-zero since T = 0 gives LO == UR)
def pwrs():
  for exp in range(2, 15):
    for mant in count(2):
      pwr = mant ** exp
      if pwr >= 10000:
        break
      s = str(pwr)
      # four digits, three different, last two equal but not zero
      if len(s) == 4 and s[2] == s[3] != '0' and len(set(s)) == 3:
        yield pwr
      
# test possible values for WATT
for watt in pwrs():
  # find the numeric digits not in WATT
  p = set(range(10)).difference(int(x) for x in str(watt))
  # now test possible values for KILO
  for q in permutations(p, 4):
    # K and L cannot be zero as they are leading digits
    if q[0] * q[2] == 0:
      continue
    kilo = 1000 * q[0] + 100 * q[1] + 10 * q[2] + q[3]
    # check that the addition result matches the pattern HOUR
    hour = kilo + watt
    if 1000 <= hour < 10000 and (hour // 100) % 10 == q[3]:
      sh = set(str(hour))
      # check that hour has four different digits, only one already known
      if len(sh) == 4 and len(sh & set(str(kilo) + str(watt))) == 1:
        fs = 'LOW = {} (KILO = {}, WATT = {}, HOUR = {})'
        print(fs.format(10 * (kilo % 100) + watt // 1000, kilo, watt, hour))

from itertools import count, permutations

# return integer powers that match the pattern WATT

# (note: T is non-zero since T = 0 gives LO == UR)

def pwrs():

for exp in range(2, 15):

for mant in count(2):

pwr = mant ** exp

if pwr >= 10000:

break

s = str(pwr)

# four digits, three different, last two equal but not zero

if len(s) == 4 and s[2] == s[3] != '0' and len(set(s)) == 3:

yield pwr

# test possible values for WATT

for watt in pwrs():

# find the numeric digits not in WATT

p = set(range(10)).difference(int(x) for x in str(watt))

# now test possible values for KILO

for q in permutations(p, 4):

# K and L cannot be zero as they are leading digits

if q[0] * q[2] == 0:

continue

kilo = 1000 * q[0] + 100 * q[1] + 10 * q[2] + q[3]

# check that the addition result matches the pattern HOUR

hour = kilo + watt

if 1000 <= hour < 10000 and (hour // 100) % 10 == q[3]:

sh = set(str(hour))

# check that hour has four different digits, only one already known

if len(sh) == 4 and len(sh & set(str(kilo) + str(watt))) == 1:

fs = 'LOW = {} (KILO = {}, WATT = {}, HOUR = {})'

print(fs.format(10 * (kilo % 100) + watt // 1000, kilo, watt, hour))

geoffrounce permalink

Taking the suggestion from Teaser 2697 on board, I permuted the 10 digits in groups of (4,3,3) digits to cut down the total number of permutations and get a reasonable run time:


from itertools import permutations

# form a reasonable list of powers
list_powers = []
for x in range(2, 50):
  for y in range(2, 25):
    pwr = x ** y
    if pwr < 10000:
        list_powers.append(pwr)

for p in permutations (range(10), 4):
  k, i, l, o = p
  if k != 0 and l != 0:
    kilo = o + l * 10 + i * 100 + k * 1000
    q = set(set(range(10)).difference(p))
    for z in permutations(q, 3):
      w, a, t = z
      if w != 0 :
        watt = t + t * 10 + a * 100 + w * 1000
        if watt in list_powers:
          j = set((q).difference(z))
          for y in permutations(j):
            h, u, r = y 
            if h != 0:
              hour = r + u * 10 + o * 100 + h * 1000
              if kilo + watt == hour:
                low = w + o * 10 + l * 100
                print('KILO WATT HOUR LOW')
                print(kilo, watt, hour, low)

from itertools import permutations

# form a reasonable list of powers

list_powers = []

for x in range(2, 50):

for y in range(2, 25):

pwr = x ** y

if pwr < 10000:

list_powers.append(pwr)

for p in permutations (range(10), 4):

k, i, l, o = p

if k != 0 and l != 0:

kilo = o + l * 10 + i * 100 + k * 1000

q = set(set(range(10)).difference(p))

for z in permutations(q, 3):

w, a, t = z

if w != 0 :

watt = t + t * 10 + a * 100 + w * 1000

if watt in list_powers:

j = set((q).difference(z))

for y in permutations(j):

h, u, r = y

if h != 0:

hour = r + u * 10 + o * 100 + h * 1000

if kilo + watt == hour:

low = w + o * 10 + l * 100

print('KILO WATT HOUR LOW')

print(kilo, watt, hour, low)

Sunday Times Teaser 2699 – Low Power

by Peter G Chamberlain

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