Sunday Times Teaser 3099 – Recurring Theme

by BRG on February 11, 2022

by Andrew Skidmore

Published Sunday February 13 2022 (link)

Liam is learning about fractions and decimals. He has been shown that some fractions give rise to decimals with a recurring pattern eg 4/11 = 0.3636… . Each pupil in his class has been given four numbers. The largest (two-figure) number is to be used as the denominator and the others as numerators, to create three fractions that are to be expressed as decimals. Liam found that each decimal was of the form 0.abcabc…, with the same digits but in different orders.

Liam’s largest number contained the digit 7 and if I told you how many of his four numbers were divisible by ten you should be able to work out the numbers.

What, in increasing order, are the four numbers?

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Brian Gladman permalink


from collections import defaultdict

# map denominators to the recurring strings 
# they produce and then map the strings when
# sorted to the numerators that produce them
d2rs = defaultdict(lambda:defaultdict(set))
for d in set(range(17, 100, 10)) | set(range(70, 80)):
  for n in range(1, d):
    # compute the recurring part using n / d = abc / 999
    abc, r = divmod(999 * n, d)
    if r == 0:
      # ensure that any leading zeros are included
      s_abc = ''.join(sorted(str(abc).zfill(3)))
      d2rs[s_abc][d].add(n)

# map the number of tens in solutions to solutions
t2s = defaultdict(set)
for abc, d_dnl in d2rs.items():
  for d, nl in d_dnl.items():
    if len(nl) >= 3:
      t4 = tuple(sorted(nl) + [d])
      t2s[sum(x % 10 == 0 for x in t4)].add(t4)

# look for a solution where the number of numbers
# divisible by ten is unique
for nt, ls in t2s.items():
  if len(ls) == 1:
    f, s, l = ', '.join(str(x) for x in ls.pop()).rpartition(', ')
    print(f"The numbers are {f} and {l}.")

from collections import defaultdict

# map denominators to the recurring strings

# they produce and then map the strings when

# sorted to the numerators that produce them

d2rs = defaultdict(lambda:defaultdict(set))

for d in set(range(17, 100, 10)) | set(range(70, 80)):

for n in range(1, d):

# compute the recurring part using n / d = abc / 999

abc, r = divmod(999 * n, d)

if r == 0:

# ensure that any leading zeros are included

s_abc = ''.join(sorted(str(abc).zfill(3)))

d2rs[s_abc][d].add(n)

# map the number of tens in solutions to solutions

t2s = defaultdict(set)

for abc, d_dnl in d2rs.items():

for d, nl in d_dnl.items():

if len(nl) >= 3:

t4 = tuple(sorted(nl) + [d])

t2s[sum(x % 10 == 0 for x in t4)].add(t4)

# look for a solution where the number of numbers

# divisible by ten is unique

for nt, ls in t2s.items():

if len(ls) == 1:

f, s, l = ', '.join(str(x) for x in ls.pop()).rpartition(', ')

print(f"The numbers are {f} and {l}.")

John Z permalink

Nested dictionaries! Witchcraft!
Below is a study I did on your solution, dialling back the sophistication to help my understanding of your code.


from collections import defaultdict

# map denominators to the recurring strings 
# they produce and then map the strings when
# sorted to the numerators that produce them
d2rs = defaultdict(set)
for d in set(range(17, 107, 10)) | set(range(70, 80)):
  for n in range(1, d):
    # find the recurring part of n/d
    abc = str(n/d)[2:5] #this might be implementation dependent

    #does it recur? does it have 3 digits?
    if abc == str(n/d)[5:8] and len(set(abc))== 3:
      s_abc = ''.join(sorted(abc))
      d2rs[s_abc + str(d)].add(n)
    
      

# map the number of tens in solutions to solutions
t2s = defaultdict(set)
for abc, nl in d2rs.items():
  
  if len(nl) >= 3:
    t4 = tuple(sorted(nl) + [int(abc[3:])])
    t2s[sum(x % 10 == 0 for x in t4)].add(t4)

# look for a solution where the number of numbers
# divisible by ten is unique
for nt, ls in t2s.items():
  if len(ls) == 1:
    f, s, l = ', '.join(str(x) for x in ls.pop()).rpartition(', ')
    print(f"The numbers are {f} and {l}.")

from collections import defaultdict

# map denominators to the recurring strings

# they produce and then map the strings when

# sorted to the numerators that produce them

d2rs = defaultdict(set)

for d in set(range(17, 107, 10)) | set(range(70, 80)):

for n in range(1, d):

# find the recurring part of n/d

abc = str(n/d)[2:5] #this might be implementation dependent

#does it recur? does it have 3 digits?

if abc == str(n/d)[5:8] and len(set(abc))== 3:

s_abc = ''.join(sorted(abc))

d2rs[s_abc + str(d)].add(n)

# map the number of tens in solutions to solutions

t2s = defaultdict(set)

for abc, nl in d2rs.items():

if len(nl) >= 3:

t4 = tuple(sorted(nl) + [int(abc[3:])])

t2s[sum(x % 10 == 0 for x in t4)].add(t4)

# look for a solution where the number of numbers

# divisible by ten is unique

for nt, ls in t2s.items():

if len(ls) == 1:

f, s, l = ', '.join(str(x) for x in ls.pop()).rpartition(', ')

print(f"The numbers are {f} and {l}.")

Sunday Times Teaser 3099 – Recurring Theme

by Andrew Skidmore

Published Sunday February 13 2022 (link)

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