Sunday Times Teaser 2656 – Wrong Adding

by Victor Bryant

In this addition, different letters consistently represent different digits but a letter in one position is wrong. The result is even.

A G A I N
+ W R O N G
===========
A D D I N G
===========

What is the correct numerical value of the total?

3 Comments Leave one →

brian gladman permalink

A bit messy this week.


from itertools import permutations

def sum_out(again, wrong, adding):
  print('  {:d}'.format(again))
  print('+ {:d}'.format(wrong))
  print('-------')
  print(' {:d}'.format(adding))
  print('-------')
  print()
  
digits = set(range(10))

# Considering the three units digits (N, G, G), if this sum is correct
# then N must be zero and this means that I is also zero from the tens
# digit sum. So the unit digit sum is incorrect. If the N is incorrect
# we again find that N = I = 0 so one of the two G's is incorrect.  So
# I = 0 with no carry or I = 9 with a carry.   Since the top digits of
# the sum are correct A must be 1 and W must be 8 or or 9. 
A = 1
for I, W in (0, 8), (0, 9), (9, 8):
  
  # permute the remaining digits for the 5 remaining letters
  for D, G, N, O, R in permutations(digits - set((I, A, W)), 5):
    
    # one of the two G's in the units position is correct so
    # either G or N + G must be even, they can't both be odd
    if G % 2 and (N + G) % 2:
      continue
    
    # find the three correct values    
    again = 10100 * A + 1000 * G + 10 * I + N
    wrong = 10000 * W + 1000 * R + 100 * O + 10 * N + G
    adding = 100000 * A + 11000 * D + 100 * I + 10 * N + G
    
    # now consider the sum without the units column and look for
    # results that are the same or differ only by a carry of one
    carry = adding // 10 - (again // 10 + wrong // 10)
    if carry in (0, 1):
      
      # if the G in ADDING is incorrect, see if there is a valid
      # digit at this position that gives a correct sum 
      dgt = (again + wrong) - (adding - G)
      if dgt % 10 in (0, 2, 4, 6, 8) and dgt // 10 == carry:
        sum_out(again, wrong, adding - G + dgt)
          
      # if the G in WRONG is incorrect, see if there is a valid
      # digit at this position that gives a correct sum
      dgt = adding - (again + wrong - G)
      if not G % 2 and dgt in range(10):
        sum_out(again, wrong - G + dgt, adding)

from itertools import permutations

def sum_out(again, wrong, adding):

print(' {:d}'.format(again))

print('+ {:d}'.format(wrong))

print('-------')

print(' {:d}'.format(adding))

print('-------')

print()

digits = set(range(10))

# Considering the three units digits (N, G, G), if this sum is correct

# then N must be zero and this means that I is also zero from the tens

# digit sum. So the unit digit sum is incorrect. If the N is incorrect

# we again find that N = I = 0 so one of the two G's is incorrect. So

# I = 0 with no carry or I = 9 with a carry. Since the top digits of

# the sum are correct A must be 1 and W must be 8 or or 9.

A = 1

for I, W in (0, 8), (0, 9), (9, 8):

# permute the remaining digits for the 5 remaining letters

for D, G, N, O, R in permutations(digits - set((I, A, W)), 5):

# one of the two G's in the units position is correct so

# either G or N + G must be even, they can't both be odd

if G % 2 and (N + G) % 2:

continue

# find the three correct values

again = 10100 * A + 1000 * G + 10 * I + N

wrong = 10000 * W + 1000 * R + 100 * O + 10 * N + G

adding = 100000 * A + 11000 * D + 100 * I + 10 * N + G

# now consider the sum without the units column and look for

# results that are the same or differ only by a carry of one

carry = adding // 10 - (again // 10 + wrong // 10)

if carry in (0, 1):

# if the G in ADDING is incorrect, see if there is a valid

# digit at this position that gives a correct sum

dgt = (again + wrong) - (adding - G)

if dgt % 10 in (0, 2, 4, 6, 8) and dgt // 10 == carry:

sum_out(again, wrong, adding - G + dgt)

# if the G in WRONG is incorrect, see if there is a valid

# digit at this position that gives a correct sum

dgt = adding - (again + wrong - G)

if not G % 2 and dgt in range(10):

sum_out(again, wrong - G + dgt, adding)

geoffrounce permalink

I get two possibilities for the ‘G’ in WRONG to be 2 or 3, (the correct value being 4 for G), leading to the same answer for ADDING:


from itertools import permutations

for p in permutations(range(10),9):
    # let k = g in wrong
    a,g,i,n,w,r,o,d,k = p
    for k in (a,g,i,n,w,o,r,d):
        if a!=0 and w != 0:
            again = n + i*10 + a*100 + g*1000 + a *10000
            wronk = k + n*10 + o*100 + r*1000 + w*10000
            adding = g + n*10 + i*100 + d*1000 + d*10000 + a*100000
            if again + wronk == adding and adding % 2 == 0:
                print('a,g,i,n,w,r,o,d,k =',p)
                print(again, '+', wronk, '=', adding)

from itertools import permutations

for p in permutations(range(10),9):

# let k = g in wrong

a,g,i,n,w,r,o,d,k = p

for k in (a,g,i,n,w,o,r,d):

if a!=0 and w != 0:

again = n + i*10 + a*100 + g*1000 + a *10000

wronk = k + n*10 + o*100 + r*1000 + w*10000

adding = g + n*10 + i*100 + d*1000 + d*10000 + a*100000

if again + wronk == adding and adding % 2 == 0:

print('a,g,i,n,w,r,o,d,k =',p)

print(again, '+', wronk, '=', adding)

brian gladman permalink

Hi Geoff,

Good to see you try this one! There are two issues here.

First, if you look at the last digit in ‘wronk’ in your output, it is not k! Why is this? The answer is that you have included k in your permutation but you then redefine it in line 6 so the k in the permutation and the k you use are different!

Second, k shouldn’t be in the permutation anyway and its value can be any digit (0 .. 9), not just those used.

Hence we get:


from itertools import permutations

for p in permutations(range(10), 8):
  a, g, i, n, w, r, o, d = p
  if a and w:
    # let G in WRONG be incorrect with the actual value k
    for k in range(10):
      again = n + i * 10 + a * 100 + g * 1000 + a * 10000
      wronk = k + n * 10 + o * 100 + r * 1000 + w * 10000
      adding = g + n * 10 + i * 100 + d * 1000 + d * 10000 + a * 100000
      if again + wronk == adding and adding % 2 == 0:
        print('(a, g, i, n, w, r, o, d, k) =', p + (k,))
        print(again, '+', wronk, '=', adding)

from itertools import permutations

for p in permutations(range(10), 8):

a, g, i, n, w, r, o, d = p

if a and w:

# let G in WRONG be incorrect with the actual value k

for k in range(10):

again = n + i * 10 + a * 100 + g * 1000 + a * 10000

wronk = k + n * 10 + o * 100 + r * 1000 + w * 10000

adding = g + n * 10 + i * 100 + d * 1000 + d * 10000 + a * 100000

if again + wronk == adding and adding % 2 == 0:

print('(a, g, i, n, w, r, o, d, k) =', p + (k,))

print(again, '+', wronk, '=', adding)

Sunday Times Teaser 2656 – Wrong Adding

by Victor Bryant

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