Sunday Times Teaser 2632 – Times Table

by Victor Bryant

I have put the numbers 1 to 16 into a four by four table in an order in which the numbers in each row increase from left to right and those in each column increase from top to bottom.

If the top row is read as one number, it has four digits and is the product of the four numbers in the second row.

What are the four numbers in the second row?

2 Comments Leave one →

brian gladman permalink

Here is my first solution:


from itertools import combinations

# the sixteen numbers used
s16 = set(range(1, 17))
# a set for storing any solutions
sol = set()

# select the first row (which are all single digits)
for p in combinations(range(1, 10), 4):
  # compute the top row as a number
  nbr = int(''.join(str(x) for x in p))

  # now select the second row from the remaining numbers
  s12 = s16 - set(p)
  for q in combinations(s12, 4):
    # check that their product equals the first row number
    if q[0] * q[1] * q[2] * q[3] == nbr:

      # check that they are greater than those above
      if all(x < y for x, y in zip(p, q)):
  
        # now set the third row from the remaining numbers
        s08 = s12 - set(q)      
        for r in combinations(s08, 4):
          
          # and check that they are greater than those above
          if all(x < y for x, y in zip(q, r)):          
            # and repeat this for the fourth row
            s = tuple(s08 - set(r))
            if all(x < y for x, y in zip(r, s)):
              # store the table for any solutions found
              sol |= set(((p, q, r, s),))

# print any solutions
for t in sol:
  print()
  for r in t:
    print('{0[0]:3d}{0[1]:3d}{0[2]:3d}{0[3]:3d}'.format(r))

from itertools import combinations

# the sixteen numbers used

s16 = set(range(1, 17))

# a set for storing any solutions

sol = set()

# select the first row (which are all single digits)

for p in combinations(range(1, 10), 4):

# compute the top row as a number

nbr = int(''.join(str(x) for x in p))

# now select the second row from the remaining numbers

s12 = s16 - set(p)

for q in combinations(s12, 4):

# check that their product equals the first row number

if q[0] * q[1] * q[2] * q[3] == nbr:

# check that they are greater than those above

if all(x < y for x, y in zip(p, q)):

# now set the third row from the remaining numbers

s08 = s12 - set(q)

for r in combinations(s08, 4):

# and check that they are greater than those above

if all(x < y for x, y in zip(q, r)):

# and repeat this for the fourth row

s = tuple(s08 - set(r))

if all(x < y for x, y in zip(r, s)):

# store the table for any solutions found

sol |= set(((p, q, r, s),))

# print any solutions

for t in sol:

print()

for r in t:

print('{0[0]:3d}{0[1]:3d}{0[2]:3d}{0[3]:3d}'.format(r))

ahmet cetinbudaklar permalink

My approach to this teaser has been to start straight away from 1234 for the top row and move on by factoring ujntil one reaches the point whereby second row carries the factors of the first row.

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Sunday Times Teaser 2632 – Times Table

by Victor Bryant

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