Sunday Times Teaser 2672 – 11, 12, 13 …

brian gladman permalink


# The lowest number (a) is divisible by 11 and at most one
# third of 2013 without repeated digits.  Since the lowest  
# number has three digits, the other two do as well.
for a in range(132, 671, 11):
  # the set of digits in a
  digits_a = set(str(a))
  if len(digits_a) == 3:
    # a larger number (b) is divisible by 12 with a + b < 2013
    for b in range(12 * ((a // 12) + 1), 2014 - a, 12):
      # the set of digits in b
      digits_b = set(str(b))
      # a and b must have six different digits between them
      if len(digits_b) == 3 and not (digits_a & digits_b):
        # the third number (c) such that a + b + c = 2013 - it
        # must be larger than a and divisible by 13 with three
        # different digits also different to those in a and b
        c = 2013 - a - b
        digits_c = set(str(c)) 
        if (c > a and c % 13 == 0 and len(digits_c) == 3 
                  and not digits_c & (digits_a | digits_b)):
          print(tuple(sorted((a, b, c))))

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

# The lowest number (a) is divisible by 11 and at most one

# third of 2013 without repeated digits. Since the lowest

# number has three digits, the other two do as well.

for a in range(132, 671, 11):

# the set of digits in a

digits_a = set(str(a))

if len(digits_a) == 3:

# a larger number (b) is divisible by 12 with a + b < 2013

for b in range(12 * ((a // 12) + 1), 2014 - a, 12):

# the set of digits in b

digits_b = set(str(b))

# a and b must have six different digits between them

if len(digits_b) == 3 and not (digits_a & digits_b):

# the third number (c) such that a + b + c = 2013 - it

# must be larger than a and divisible by 13 with three

# different digits also different to those in a and b

c = 2013 - a - b

digits_c = set(str(c))

if (c > a and c % 13 == 0 and len(digits_c) == 3

and not digits_c & (digits_a | digits_b)):

print(tuple(sorted((a, b, c))))

Reply

geoffrounce permalink

Here is a slower permutation solution:


from itertools import permutations

for p in permutations((1,2,3,4,5,6,7,8,9,0),9):
    a,b,c,d,e,f,g,h,i = p
    if all(x != 0 for x in(a,d,g)):
        n1 = c + b*10 + a*100
        n2 = f + e*10 + d*100
        n3 = i + h*10 + g*100
        m = sorted(set((n1,n2,n3)))
        if m[0]<m[1] and m[1]<m[2] and m[0]+m[1]+m[2] == 2013:
            if (m[0] % 11 == 0 and m[1] % 12 == 0 and m[2] % 13 == 0) 
               or (m[0] % 11 == 0 and m[1] % 13 == 0 and m[2] % 12 == 0):
                print(m[0],m[1],m[2])

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

from itertools import permutations

for p in permutations((1,2,3,4,5,6,7,8,9,0),9):

a,b,c,d,e,f,g,h,i = p

if all(x != 0 for x in(a,d,g)):

n1 = c + b*10 + a*100

n2 = f + e*10 + d*100

n3 = i + h*10 + g*100

m = sorted(set((n1,n2,n3)))

if m[0]<m[1] and m[1]<m[2] and m[0]+m[1]+m[2] == 2013:

if (m[0] % 11 == 0 and m[1] % 12 == 0 and m[2] % 13 == 0)

or (m[0] % 11 == 0 and m[1] % 13 == 0 and m[2] % 12 == 0):

print(m[0],m[1],m[2])

Reply

ahmet cetinbudaklar permalink

Taking the three numbers to be a 3-digit number, the smallest one cannot be larger than 594 since the sum of the three numbers must be equal to 2013 and the first digit of the other two numbers can be 6,7,8 or even 9.
Thus we can approach the etaser by trial and error as seen from the table below:

/11 /12 /13
594 612 807x
583 672 x
572 648 x
561 708 x
539 840 x
528 924 x
517 804 692x
506 732 x
495 816 702 which gives us the solution.

Reply

brian gladman permalink

Hi Geoff,

You might be interested to know that a permutation based solution can be made very fast by handling the units, ten and hundreds columns of the sum separately as follows:


from itertools import combinations, permutations, product

# The sum uses nine of the ten digits - using digital roots (dr) we have
# dr(0123456789) = dr(2013) + dr(missing digit)  - hence '3' is missing
d0 = set((0, 1, 2, 4, 5, 6, 7, 8, 9))
# select three digits for the units column
for u in combinations(d0, 3):
  # the carry and the column sum (which must be 3)
  ct, su = divmod(sum(u), 10)
  if su == 3:
    # the remaining digits for tens and hundreds columns
    d1 = d0 - set(u)
    # select three digits for the tens column
    for t in combinations(d1, 3):
      # the carry and the column sum (which must be 1)
      ch, st = divmod(sum(t) + ct, 10)
      if st == 1:
        # the remaining digits for the hundreds column        
        d2 = d1 - set(t)
        # select digits for the hundreds column
        for h in combinations(d2, 3):
          # the sum of the hundreds column with carry must be twenty
          if sum(h) + ch == 20:
            # now permutate the digits in the tens and hundreds columns
            for p, q in product(*(permutations(c) for c in (t, h))):
              # form the three numbers and check that the lowest is a multiple
              # of 11 and that the others are multiples of 12 and 13
              a, b, c = sorted(100 * x + 10 * y + z for x,y,z in zip(q, p, u))
              if not (a % 11 or (b % 12 or c % 13) and (b % 13 or c % 12)):
                print('The numbers are {}, {} and {}.'.format(a, b, c))

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

from itertools import combinations, permutations, product

# The sum uses nine of the ten digits - using digital roots (dr) we have

# dr(0123456789) = dr(2013) + dr(missing digit) - hence '3' is missing

d0 = set((0, 1, 2, 4, 5, 6, 7, 8, 9))

# select three digits for the units column

for u in combinations(d0, 3):

# the carry and the column sum (which must be 3)

ct, su = divmod(sum(u), 10)

if su == 3:

# the remaining digits for tens and hundreds columns

d1 = d0 - set(u)

# select three digits for the tens column

for t in combinations(d1, 3):

# the carry and the column sum (which must be 1)

ch, st = divmod(sum(t) + ct, 10)

if st == 1:

# the remaining digits for the hundreds column

d2 = d1 - set(t)

# select digits for the hundreds column

for h in combinations(d2, 3):

# the sum of the hundreds column with carry must be twenty

if sum(h) + ch == 20:

# now permutate the digits in the tens and hundreds columns

for p, q in product(*(permutations(c) for c in (t, h))):

# form the three numbers and check that the lowest is a multiple

# of 11 and that the others are multiples of 12 and 13

a, b, c = sorted(100 * x + 10 * y + z for x,y,z in zip(q, p, u))

if not (a % 11 or (b % 12 or c % 13) and (b % 13 or c % 12)):

print('The numbers are {}, {} and {}.'.format(a, b, c))

This is significantly faster than my first solution above.

Reply

Sunday Times Teaser 2672 – 11, 12, 13 …

by Victor Bryant

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta