Sunday Times Teaser 2664 – Time ‘n’ Again

brian gladman permalink


from itertools import permutations

# Since the fraction on the left:
#
#   TIME     AGAIN
#  -----  =  -----
#  **N**     99999
#
# is in reduced form with a five digit denominator, the
# latter must be 11111, 33333 or 99999. But it can't be
# 99999 since TIME would then equal AGAIN.
for N in '13':
  d = { 'N':N }

  # now check possible leading digits for A and T
  for p in permutations(set('123456789') - set((N,)), 2):
    d.update(zip('AT', p))

    # now check possible digits for G, I, M and E
    for q in permutations(set('0123456789') - set(p + (N,)), 4):
      d.update(zip('GIME', q))

      # now form AGAIN and check that it is divisible by (9 / n)
      # and that this division gives TIME 
      AGAIN = int(''.join(d[c] for c in 'AGAIN'))
      time, r = divmod(AGAIN, 9 // int(N))
      if not r and time == int(''.join(d[c] for c in 'TIME')):
        print('TIME = {:d} (AGAIN = {:d})'.format(time, AGAIN))

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from itertools import permutations

# Since the fraction on the left:

#

# TIME AGAIN

# ----- = -----

# **N** 99999

#

# is in reduced form with a five digit denominator, the

# latter must be 11111, 33333 or 99999. But it can't be

# 99999 since TIME would then equal AGAIN.

for N in '13':

d = { 'N':N }

# now check possible leading digits for A and T

for p in permutations(set('123456789') - set((N,)), 2):

d.update(zip('AT', p))

# now check possible digits for G, I, M and E

for q in permutations(set('0123456789') - set(p + (N,)), 4):

d.update(zip('GIME', q))

# now form AGAIN and check that it is divisible by (9 / n)

# and that this division gives TIME

AGAIN = int(''.join(d[c] for c in 'AGAIN'))

time, r = divmod(AGAIN, 9 // int(N))

if not r and time == int(''.join(d[c] for c in 'TIME')):

print('TIME = {:d} (AGAIN = {:d})'.format(time, AGAIN))

Reply

brian gladman permalink

Another version that is much faster along the lines of Arthur’s below:


# Since the fraction on the left:
#
#   TIME     AGAIN
#  -----  =  -----
#  **N**     99999
#
# is in reduced form with a five digit denominator, the
# latter must be 11111, 33333 or 99999. But it can't be
# 99999 since TIME would then equal AGAIN.  So N can be
# 1 or 3 and AGAIN = (9 / N) * TIME
for N in (1, 3):
  for time in range(1000, 10000):
    # t is 'time' as a string
    t = str(time)
    if len(set(t)) == 4:
      # a is 'again' as a string
      a = str((9 // N) * time)
      if (len(a) == 5 and a[0] == a[2] and a[4] == str(N)
          and t[1] == a[3] and len(set(t + a)) == 7):
        print('TIME = {:s} (AGAIN = {:s})'.format(t, a))

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# Since the fraction on the left:

#

# TIME AGAIN

# ----- = -----

# **N** 99999

#

# is in reduced form with a five digit denominator, the

# latter must be 11111, 33333 or 99999. But it can't be

# 99999 since TIME would then equal AGAIN. So N can be

# 1 or 3 and AGAIN = (9 / N) * TIME

for N in (1, 3):

for time in range(1000, 10000):

# t is 'time' as a string

t = str(time)

if len(set(t)) == 4:

# a is 'again' as a string

a = str((9 // N) * time)

if (len(a) == 5 and a[0] == a[2] and a[4] == str(N)

and t[1] == a[3] and len(set(t + a)) == 7):

print('TIME = {:s} (AGAIN = {:s})'.format(t, a))

Reply

BRG permalink

More analysis and even faster


from itertools import permutations

# Since the fraction on the left:
#
#   TIME     AGAIN
#  -----  =  -----
#  **N**     99999
#
# is in reduced form with a five digit denominator, the
# latter must be 11111, 33333 or 99999. But it can't be
# 99999 since TIME would then equal AGAIN. Hence:
#
#     99999.TIME = 11111.N.AGAIN
#
# which can be expressed in thne form:
#
#     9 x TIME = [N + 5.AGAI]^2 - [5.AGAI]^2

for N in (1, 3):
  for A, G, I in permutations(set(range(10)) - {N}, 3):
    AGAI = 1010 * A + 100 * G + I
    TIME, r = divmod((N + 5 * AGAI) ** 2 - (5 * AGAI) ** 2, 9)
    t = str(TIME)
    if not r and len(t) == len(set(t)) == 4 and str(I) == t[1]:
      T, _, M, E = (int(c) for c in t)
      s7 = (N, A, G, I, T, M, E)
      if len(s7) == len(set(s7)) == 7:
        print(f"{TIME = } (AGAIN = {10 * AGAI + N})")

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from itertools import permutations

# Since the fraction on the left:

#

# TIME AGAIN

# ----- = -----

# **N** 99999

#

# is in reduced form with a five digit denominator, the

# latter must be 11111, 33333 or 99999. But it can't be

# 99999 since TIME would then equal AGAIN. Hence:

#

# 99999.TIME = 11111.N.AGAIN

#

# which can be expressed in thne form:

#

# 9 x TIME = [N + 5.AGAI]^2 - [5.AGAI]^2

for N in (1, 3):

for A, G, I in permutations(set(range(10)) - {N}, 3):

AGAI = 1010 * A + 100 * G + I

TIME, r = divmod((N + 5 * AGAI) ** 2 - (5 * AGAI) ** 2, 9)

t = str(TIME)

if not r and len(t) == len(set(t)) == 4 and str(I) == t[1]:

T, _, M, E = (int(c) for c in t)

s7 = (N, A, G, I, T, M, E)

if len(s7) == len(set(s7)) == 7:

print(f"{TIME = } (AGAIN = {10 * AGAI + N})")

Reply

Frits permalink


from itertools import permutations
 
# see former analysis
#
#     9 x TIME = N.AGAIN
#
# if N = 3 then E = 1 and A = 2 (A < 4)
 
for N in (1, 3):
  # if N = 1 then E must be 9
  mx = 8 if N == 1 else 9
  for G, I in permutations(set(range(mx + 1)) - {1, N}, 2):
    # AGAIN must be a ninefold or a threefold
    r = 9 - (NGI := N + G + I) % 9
    A = (r + 9 * (r % 2)) // 2 if N == 1 else 2
    # 9.xxxx can never become 9xxxx
    if A in {N, G, I, 9}: continue
    if N == 3 and (NGI + 2 * A) % 3: continue 
    
    AGAIN = 10100 * A + 1000 * G + 10 * I + N
    TIME = N * AGAIN // 9
    #print("N/G/I", (N, G, I), "A =", A, N * AGAIN / 9)
    if not (999 < TIME < 10000): continue
    T, I_, M, E = (int(c) for c in str(TIME))
    if I != I_: continue
    if len({N, A, G, T, I, M, E}) != 7: continue
    print(f"{TIME=}")

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from itertools import permutations

# see former analysis

#

# 9 x TIME = N.AGAIN

#

# if N = 3 then E = 1 and A = 2 (A < 4)

for N in (1, 3):

# if N = 1 then E must be 9

mx = 8 if N == 1 else 9

for G, I in permutations(set(range(mx + 1)) - {1, N}, 2):

# AGAIN must be a ninefold or a threefold

r = 9 - (NGI := N + G + I) % 9

A = (r + 9 * (r % 2)) // 2 if N == 1 else 2

# 9.xxxx can never become 9xxxx

if A in {N, G, I, 9}: continue

if N == 3 and (NGI + 2 * A) % 3: continue

AGAIN = 10100 * A + 1000 * G + 10 * I + N

TIME = N * AGAIN // 9

#print("N/G/I", (N, G, I), "A =", A, N * AGAIN / 9)

if not (999 < TIME < 10000): continue

T, I_, M, E = (int(c) for c in str(TIME))

if I != I_: continue

if len({N, A, G, T, I, M, E}) != 7: continue

print(f"{TIME=}")

Reply

Arthur Vause permalink

Similar analysis to Brian, using the factorisation of 99999=3^2*41*271, but using the equation

3^2*41*271 * TIME = AGAIN * xxNxx, and distributing the factors of 99999 amongst AGAIN and xxNxx


from fractions import gcd

for d in (3,41,271):
  for n in xrange(10000/d+1, 100000/d):
    again=str(n*d)
    if again[0]==again[2] and len(set(again))==4:
      m = 99999 / gcd(int(again),99999)
      for x in xrange(10000/m+1,100000/m):
        xxnxx = x*m
        if str(xxnxx)[2]==again[4]:
          time = (int(again)*xxnxx)/99999
          if 1000<=time<=9999 and str(time)[1]==again[3]:
            if len(set(str(time)+str(again)))==7:
              if gcd(xxnxx,time)==1 :
                print time, "/", xxnxx,"=0.(",again,")"

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from fractions import gcd

for d in (3,41,271):

for n in xrange(10000/d+1, 100000/d):

again=str(n*d)

if again[0]==again[2] and len(set(again))==4:

m = 99999 / gcd(int(again),99999)

for x in xrange(10000/m+1,100000/m):

xxnxx = x*m

if str(xxnxx)[2]==again[4]:

time = (int(again)*xxnxx)/99999

if 1000<=time<=9999 and str(time)[1]==again[3]:

if len(set(str(time)+str(again)))==7:

if gcd(xxnxx,time)==1 :

print time, "/", xxnxx,"=0.(",again,")"

Reply

Arthur Vause permalink

And a faster version, using Brian’s observation that **N** has to be 11111 or 33333


from fractions import gcd
 
for xxnxx in (11111, 33333):
  n = str(xxnxx)[2]
  for time in range(1234, 9876):
    if gcd(time, xxnxx) == 1:
      again = str(time * 99999 // xxnxx)
      if (len(again) == 5
          and again[0] == again[2]
          and n == again[4]
          and str(time)[1] == again[3]
          and len(set(str(time) + str(again))) == 7):
        print(time, "/", xxnxx,"= 0.(",again,")")

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from fractions import gcd

for xxnxx in (11111, 33333):

n = str(xxnxx)[2]

for time in range(1234, 9876):

if gcd(time, xxnxx) == 1:

again = str(time * 99999 // xxnxx)

if (len(again) == 5

and again[0] == again[2]

and n == again[4]

and str(time)[1] == again[3]

and len(set(str(time) + str(again))) == 7):

print(time, "/", xxnxx,"= 0.(",again,")")

Reply

brian gladman permalink

Hi Arthur

I hope you don’t mind but I have deleted the incorrect version and, while I was at it, I converted your corrected version to Python 3 (I get emails telling me that some code here doesn’t work and I often find that this is because people are using Python 3 without even realising that there is a Python 2).

We seem to have converged on very similar solutions 🙂

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Sunday Times Teaser 2664 – Time ‘n’ Again

by Victor Bryant

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