Twisted Addition Cryptogram!

Computer Science Level 4

$\large{\begin{array}{ccccccc} && & & & T & E&N\\ && & & & T & E&N\\ && & & N & I &N&E \\ && & E&I &G &H&T\\ + && & T&H &R & E&E \\ \hline & & &F & O&R & T&Y\\ \hline \end{array}}$

How many solutions exists to the given addition problem, where each letter represents a distinct digit?

Note: The first digit of a number can be zero. For example (not necessarily true), in the number $\overline{NINE}$ , we can have it as $\overline{0I0E}$ .

2 10 8 5 12 6 0 4

3 solutions

Satyajit Mohanty
Aug 7, 2015

There are 10 solutions:

$417 + 417 + 7871 + 18324 + 42511 = 69540$

$701 + 701 + 1410 + 04267 + 76500 = 83579$

$047 + 047 + 7174 + 41690 + 09344 = 58302$

$718 + 718 + 8281 + 12347 + 74011 = 96075$

$148 + 148 + 8284 + 42391 + 19644 = 70615$

$713 + 713 + 3431 + 14207 + 70611 = 89675$

$631 + 631 + 1013 + 30756 + 65233 = 98264$

$329 + 329 + 9492 + 24863 + 36022 = 71035$

$624 + 624 + 4042 + 20756 + 65322 = 91368$

$308 + 308 + 8780 + 07643 + 34200 = 51239$

Moderator note:

How can we show that there are only 10 solutions?

You made a list of some solutions for the problem, but you have to prove that there are only 10 solutions.

Paulo Guilherme Santos - 5 years, 10 months ago

Anonymous1 Assassin
Jan 23, 2021

I have written the shortest and the most concise code I can get for solving this problem:

l=0
for t,e,n,i,g,h,r,f,o,y in permutations('0123456789',10):
  if 2*int(t+e+n)+int(n+i+n+e)+int(e+i+g+h+t)+int(t+h+r+e+e)==int(f+o+r+t+y):
    l+=1
print(l)

1 2	`Output: 10`

Soumava Pal
Mar 19, 2016

for t in range(10):
    for e in range(10):
        if e!=t:
            for n in range(10):
                if n not in [e,t]:
                    for i in range(10):
                        if i not in [t,e,n]:
                            for g in range(10):
                                if g not in [t,e,n,i]:
                                    for h in range(10):
                                        if h not in [t,e,n,i,g]:
                                            for r in range(10):
                                                if r not in [t,e,n,i,g,h]:
                                                    for f in range(10):
                                                        if f not in [t,e,n,i,g,h,r]:
                                                            for o in range(10):
                                                                if o not in [t,e,n,i,g,h,r,f]:
                                                                    for y in range(10):
                                                                        if y not in [t,e,n,i,g,h,r,f,o]:
                                                                            if (200*t+20*e+2*n+1000*n+100*i+10*n+e+10000*e+1000*i+100*g+10*h+t+10000*t+1000*h+100*r+10*e+e)==(10000*f+1000*o+100*r+10*t+y):
                                                                                print t,e,n,i,g,h,r,f,o,y

0 4 7 1 6 9 3 5 8 2
1 4 8 2 3 9 6 7 0 5
3 0 8 7 6 4 2 5 1 9
3 2 9 4 8 6 0 7 1 5
4 1 7 8 3 2 5 6 9 0
6 2 4 0 7 5 3 9 1 8
6 3 1 0 7 5 2 9 8 4
7 0 1 4 2 6 5 8 3 9
7 1 3 4 2 0 6 8 9 5
7 1 8 2 3 4 0 9 6 5
>>>

This is pretty good and straightforward but we can use permutations from itertools to shorten it up a LiTtLe BiT lol:

l=0
for t,e,n,i,g,h,r,f,o,y in permutations('0123456789',10):
  if 2*int(t+e+n)+int(n+i+n+e)+int(e+i+g+h+t)+int(t+h+r+e+e)==int(f+o+r+t+y):
    l+=1
print(l)

1 2	`Output: 10`

Anonymous1 Assassin - 4 months, 3 weeks ago

Twisted Addition Cryptogram!

3 solutions

Moderator note:

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