Sunday Times Teaser 2611 – Simple Arithmetic

by Danny Roth

Published: 7 October 2012 (link)

George and Martha are teaching their great-grandchildren some simple arithmetic. “If you add two thirties to four tens you get a hundred,” George was saying, and he wrote it like this:

“Strangely,” added Martha, there are nine different letters there, and if you allow each letter to stand for a different digit, there is a unique sum which works.”

Which digit would be missing?

4 Comments Leave one →

Brian Gladman permalink

Here is my solution:


from itertools import permutations
from collections import Counter

digits = '0123456789'
sol = Counter()
for p in permutations(digits, 9):
  # skip any permutations giving leading zeroes
  if '0' in p[:2]:
    continue
  # map letters to the digits they represent
  d = dict(zip('THDEINRUY', p))

  # compute the numeric values for this permutation
  thirty = int(''.join(d[c] for c in 'THIRTY'))
  ten = int(''.join(d[c] for c in 'TEN'))
  hundred = int(''.join(d[c] for c in 'HUNDRED'))

  # count the correct sums for each missing digit
  if 2 * thirty + 4 * ten == hundred:
    sol[min(set(digits) - set(p))] += 1

# ouput the missing digit for any unique sums
for c in sol:
  if sol[c] == 1:
    print('The missing digit is {:s}.'.format(c))

from itertools import permutations

from collections import Counter

digits = '0123456789'

sol = Counter()

for p in permutations(digits, 9):

# skip any permutations giving leading zeroes

if '0' in p[:2]:

continue

# map letters to the digits they represent

d = dict(zip('THDEINRUY', p))

# compute the numeric values for this permutation

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

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

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

# count the correct sums for each missing digit

if 2 * thirty + 4 * ten == hundred:

sol[min(set(digits) - set(p))] += 1

# ouput the missing digit for any unique sums

for c in sol:

if sol[c] == 1:

print('The missing digit is {:s}.'.format(c))

geoffrounce permalink

A different approach using permutations:


from itertools import permutations
for p in permutations('0123456789', 9):
  t = int(p[0])
  if t != 0:
    h = int(p[1])
    if h != 0:
      i = int(p[2]); r = int(p[3]); y = int(p[4])
      e = int(p[5]); n = int(p[6]); u = int(p[7])
      d = int(p[8])
      # now construct the equation
      thirty = t*100000 + h*10000 + i*1000 + r*100 + 10*t +y
      ten = t*100 + e*10 + n
      hundred = h*1000000 + u*100000 + n*10000 + d*1000 + r*100 + e*10 + d
      if 2*thirty + 4*ten == hundred:
        print('thirty,ten, hundred = ',thirty,ten, hundred)
        print('t,h,i,r,y,e,n,u,d = ',t,h,i,r,y,e,n,u,d) # ie 915260384
        # gives 2 * 915,296 + 4 * 903 = 1,834,204
        # so missing digit = 7

from itertools import permutations

for p in permutations('0123456789', 9):

t = int(p[0])

if t != 0:

h = int(p[1])

if h != 0:

i = int(p[2]); r = int(p[3]); y = int(p[4])

e = int(p[5]); n = int(p[6]); u = int(p[7])

d = int(p[8])

# now construct the equation

thirty = t*100000 + h*10000 + i*1000 + r*100 + 10*t +y

ten = t*100 + e*10 + n

hundred = h*1000000 + u*100000 + n*10000 + d*1000 + r*100 + e*10 + d

if 2*thirty + 4*ten == hundred:

print('thirty,ten, hundred = ',thirty,ten, hundred)

print('t,h,i,r,y,e,n,u,d = ',t,h,i,r,y,e,n,u,d) # ie 915260384

# gives 2 * 915,296 + 4 * 903 = 1,834,204

# so missing digit = 7

Brian Gladman permalink

Here is a further development of Geoff’s solution that is very fast on pypy:


from itertools import permutations
from collections import Counter

digits = range(10)
# a counter for solutions for each unused digit
sol = Counter()
# permute the ten digits, nine at a time
for p in permutations(digits, 9):
  # ensure no zero leading digits
  if p[0] and p[1]:
    # allocate digits to letters
    t, h, i, r, y, e, n, u, d = p

    # compile the numeric values for the addition
    thirty =  10 * (10 * (10 * (10 * (10 * t + h) + i) + r) + t) + y
    ten = 10 * (10 * t + e) + n
    hundred = 10 * (10 * (10 * (10 * (10 * (10 * h + u) + n) + d) + r) + e) + d

    # count the number of correct solutions for each missing digit
    if 2 * thirty + 4 * ten == hundred:
      c = min(set(digits) - set(p))
      sol[c] += 1

# output the missing character for any unique solutions
for c in sol:
  if sol[c] == 1:
    print('Missing digit is {:d}.'.format(c))

from itertools import permutations

from collections import Counter

digits = range(10)

# a counter for solutions for each unused digit

sol = Counter()

# permute the ten digits, nine at a time

for p in permutations(digits, 9):

# ensure no zero leading digits

if p[0] and p[1]:

# allocate digits to letters

t, h, i, r, y, e, n, u, d = p

# compile the numeric values for the addition

thirty = 10 * (10 * (10 * (10 * (10 * t + h) + i) + r) + t) + y

ten = 10 * (10 * t + e) + n

hundred = 10 * (10 * (10 * (10 * (10 * (10 * h + u) + n) + d) + r) + e) + d

# count the number of correct solutions for each missing digit

if 2 * thirty + 4 * ten == hundred:

c = min(set(digits) - set(p))

sol[c] += 1

# output the missing character for any unique solutions

for c in sol:

if sol[c] == 1:

print('Missing digit is {:d}.'.format(c))

geoffrounce permalink

As part of the joint effort, I think we should give some credit to Naim. I had seen him permute individual digits from letters to make larger numbers on Jim Randall’s Enigma site. I thought it was a method worth trying for this teaser.

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Sunday Times Teaser 2611 – Simple Arithmetic

by Danny Roth

Published: 7 October 2012 (link)

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