New Scientist Enigma 631 – Fidget Digit

by BRG on July 6, 2021

by Susan Denham

From Issue #1785, 7th September 1991 (link)

My niece (whose age is a two-figure number) is very good at arithmetic, and to keep her occupied on a recent train journey I asked her to find a number with all its digits different and with the sum of its digits a multiple (more than one times) of her age.

She wrote down a list of lots of number with both those properties. So I then asked her to add 1 to each of her numbers, and to pick out from the new list each number which still had all its digits different and with the sum of its digits equal to a multiple (again, more than one times) of her age next birthday.

There were still quite a few numbers in this new list which had these properties. So I asked her to find one of them which, when multiplied by her age, gave an answer which still had all its digits different — which she did!

How old is she?

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3 Comments Leave one →

Brian Gladman permalink


from itertools import combinations, permutations
from collections import defaultdict
from functools import reduce

age_to_dgts = defaultdict(set)
# the length of the numbers found
for nd in range(3, 11):
  # consider the digits in the numbers found
  for dgts in combinations(range(10), nd):
    dgt_sum = sum(dgts)
    # the digit sum (<= 45) is a multiple (> 1) of the
    # two digit age; hence consider possible multiples
    if dgt_sum >= 20:
      for mlt in (2, 3, 4):
        age, r = divmod(dgt_sum, mlt)
        if not r and age >= 10:
          # consider the last two digits of possible 
          # numbers with these digits
          for p, l in permutations(dgts, 2):
            # the other digits
            ndgts = tuple(d for d in dgts if d not in {p, l})
            # find the last two digits after adding 1 to the number
            if l < 9:
              ndgts += (p, l + 1)
            else:
              ndgts += (p + 1, 0)
            # the digits must all still be different ...
            if len(ndgts) == len(set(ndgts)):
              # ... and have a digit sum that is a multiple 
              # of the age at the next birthday 
              m, r = divmod(sum(ndgts), age + 1)
              if not r and m > 1:
                # save the digits of the new number
                age_to_dgts[age].add(ndgts)

# consider the possible ages and sets of 
# digits in the resulting list of numbers
for age, l_dgts in age_to_dgts.items():
  nbrs = []
  # consider the digits whose positions have not been fixed
  # and the last two digits with fixed positions and values
  for *dgts, p, l in l_dgts:
    # permute the digits whose positions are not yet fixed
    for q in permutations(dgts):
      # the resulting number
      nbr = reduce(lambda x, y: 10 * x + y, q + (p, l))
      # .... when multiplied by the age gives a result 
      # in which all the digits are different
      s = str(age * nbr)
      if len(s) == len(set(s)):
        nbrs.append(nbr)
  
  if nbrs:
    print(f"age = {age} ==> {len(nbrs)} final numbers:")
    for i, n in enumerate(nbrs, 1):
      print(f"{i:>2d}  {n} ==> {age * n}")

from itertools import combinations, permutations

from collections import defaultdict

from functools import reduce

age_to_dgts = defaultdict(set)

# the length of the numbers found

for nd in range(3, 11):

# consider the digits in the numbers found

for dgts in combinations(range(10), nd):

dgt_sum = sum(dgts)

# the digit sum (<= 45) is a multiple (> 1) of the

# two digit age; hence consider possible multiples

if dgt_sum >= 20:

for mlt in (2, 3, 4):

age, r = divmod(dgt_sum, mlt)

if not r and age >= 10:

# consider the last two digits of possible

# numbers with these digits

for p, l in permutations(dgts, 2):

# the other digits

ndgts = tuple(d for d in dgts if d not in {p, l})

# find the last two digits after adding 1 to the number

if l < 9:

ndgts += (p, l + 1)

else:

ndgts += (p + 1, 0)

# the digits must all still be different ...

if len(ndgts) == len(set(ndgts)):

# ... and have a digit sum that is a multiple

# of the age at the next birthday

m, r = divmod(sum(ndgts), age + 1)

if not r and m > 1:

# save the digits of the new number

age_to_dgts[age].add(ndgts)

# consider the possible ages and sets of

# digits in the resulting list of numbers

for age, l_dgts in age_to_dgts.items():

nbrs = []

# consider the digits whose positions have not been fixed

# and the last two digits with fixed positions and values

for *dgts, p, l in l_dgts:

# permute the digits whose positions are not yet fixed

for q in permutations(dgts):

# the resulting number

nbr = reduce(lambda x, y: 10 * x + y, q + (p, l))

# .... when multiplied by the age gives a result

# in which all the digits are different

s = str(age * nbr)

if len(s) == len(set(s)):

nbrs.append(nbr)

if nbrs:

print(f"age = {age} ==> {len(nbrs)} final numbers:")

for i, n in enumerate(nbrs, 1):

print(f"{i:>2d} {n} ==> {age * n}")

Frits permalink

@Brian, if l is equal to 9 and age ends on a zero you can discard these entries (otherwise age times second list number will end on 2 zeros).

Reply
- Brian Gladman permalink
  
  @Frits
  
  Thanks for the thought. I would add this as a special case if the speed was
  a problem but it isn’t. I don’t generally add special cases unless they are
  needed because a program is unduly slow without them. Adding them often
  makes it necessary to add a long comment to explain them and this makes the
  program itself longer. Of course a single special case is not a big problem
  but it is easy when speed is made a priority to end up adding multiple
  special cases which have a negative impact on the overall clarity of the code.
  
  Reply

New Scientist Enigma 631 – Fidget Digit

by Susan Denham

From Issue #1785, 7th September 1991 (link)

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