0
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Similar to the previous puzzle, find the values behind the letters.

     T , E , M , A 
     K , A , K , A 
+    S , A , F , T 
-------------------
 F , E , S , T , A  

These are roughly the Swedish words for theme, cookie, soda, and the result, partying. All letters correspond to different digits, and leading digits in numbers are non-zero.

Below is a second puzzle of this type:

    A , N , A , N , A , S 
    M , A , T , O , S , T 
        T , O , M , A , T 
+           S , A , L , T 
-------------------------------
    S , A , L , L , A , D 

The words are pineapple, cheese, tomato, salt with the total salad.

The solutions are unique.

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1
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Human unfriendly way to solve first puzzle.

Let we call each carry digits c1 to c4.

We have:
[0]: 0 <= c1, c2, c3, c4 <= 2
[1]: A+A+T=c1*10+A
[2]: M+K+F+c1=T+c2*10
[3]: E+A+A+c2=S+c3*10
[4]: T+K+S+c3=E+c4*10; T<>0; K<>0; S<>0
[5]: F=c4; F<>0

----

[1] -> A+T=c1*10
-> [6]: c1=1;
-> [7]: A+T=10; A<>0; T<>0

---

[4] + [3] -> T+K+(E+A+A+c2-c3*10)+c3=E+c4*10
-> T+K+A+A+c2=c4*10+c3*9
+ [2] -> (M+K+F+1-c2*10)+K+A+A+c2=c4*10+c3*9
-> [8]: M+K+K+A+A+1=(c4+c2+c3)*9

[9]: 12 = 4+1+1+3+3 <= M+K+K+A+A <= 7+8+8+9+9 = 41
+ [8] -> 12 <= 9*(c2+c3+c4) <= 42
-> [10]: 2<=c2+c3+c4<=4

[4] + [6] + [2] -> (M+K+F+1-c2*10)+K+S+c3=E+c4*10
+ [5] -> [11]: M+K+K+S+1+c3+c4-E=(c2+c4)*10

K<>F=c4; K<>0
M+K+K+S+1+c3+c4-E<=7+8+9+9+1+2+2-0=38
-> (c2+c4)*10<=37
-> [12]: c2+c4<=3

----

Consider c2=2;
[2]: M+K+F+1=T+c2*10>=1+2*10=21
-> M+K+F>=20
-> M+K>=20-F
+ [0] -> M+K>=18
M, K cannot both be 9 (impossible)
-> [13]: c2<>2

----

-> [8]: M+K+K+A+A+1=(c4+c2+c3)*9
-> [7]: A+T=10; A<>0; T<>0
[2]: M+K+F+1=T+c2*10

[2] + [8] + [7] -> M+K=(10-A)+c2*10-1-c4=10+c2*10-A-1-c4
-> M+K+K=9*c2+9*c3+9*c4-1-A-A
-> [14]: A+K=9*c3+10*c4-c2-10
-> [15]: 14<=9c3+10c4-c2<=27

----

[2] + [3] + [4]
-> (E+A+A+c2)+(T+K+S+c3)+(T+c2*10)=(M+K+F+c1)+(S+c3*10)+(E+c4*10)
-> c2*11+20=M+F+1+c3*9+c4*10
-> [16]: M=19+c2*11-c3*9-c4*11
-> 10<=-c2*11+c3*9+c4*11<=19
-> 10<=11*(c3+c4-c2)-2*c3<=19
-> 10<=11*(c3+c4-c2)<=23
-> c3+c4-c2=1 or (c3=2 and c2=c4=1)
If (c3=2, c2=c4=1) -> M=1=F (impossible)
-> c3+c4-c2=1

---

[10] + [12] + [13] + [15] + [17]
-> [17]: (c2, c3, c4) may be (1, 0, 2) or (1, 1, 1)
-> [18]: c2+c3+c4=3
-> [19]: c2=1
-> [20]: c3+c4=2

+ [8]
-> [21]: M+K+K+A+A=26

[14] -> [22]: A+K=7+c4
[16] -> [23]: M+F=12-c4

[21] + [22]
-> M=12-c4*2
-> [24]: F=c4=2
-> [25]: M=8
-> [26]: c3=0
-> [27]: A+K=9; K=T-1

[3] -> [28]: E+A+A+1=S

[27] + [7] + [28] + [25] + [24]:
(A, T, K, E, S, M, F) = (4, 6, 5, 0, 9, 8, 2)

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0
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N = 0, A=1, L = 2, M = 3, S = 5, O = 7, T = 8, D= 9

101015

+ 318758

+ 87318

+ 5128

= 512219

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0
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Attempting a Human friendly way to solve as requested by @Per Alexandersson. Work in progress: Puzzle 1:

F can only be 1 or 2 Reason: overflow digit of addition of 3 numbers.

considering @tsh suggestion

A + T = 10 Reason: A+A+T = A , so A+T should be 10 and the answer will be A+A+T = 1A (1 carried over).

Which implies:

1. A, T cannot be 0 , since if A = 0 , T = 10 vice versa. 2. A, T cannot be 5. Since A+T = 10 , A = 5 then T = 5. But A,T are different numbers(i hope).

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  • 1
    $\begingroup$ "A and T are either both odd or both even" -> Why not just A+T=10. $\endgroup$ – tsh Nov 22 '20 at 12:25
  • $\begingroup$ @tsh Oh i realise you are right, A+T = 10 and neither A nor T can be 5. Does that sound better ?! $\endgroup$ – Vipul Tawde Nov 22 '20 at 12:30
0
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From my previous answer, we can find the answers with brute force:

from itertools import permutations

for comb in permutations([1, 2, 3, 4, 5, 6, 7, 8, 9, 0], 7):
    a, e, f, k, m, s, t = comb
    if all([t, k, s, f]):
        if int(f'{t}{e}{m}{a}') + int(f'{k}{a}{k}{a}') + int(f'{s}{a}{f}{t}') == int(f'{f}{e}{s}{t}{a}'):
            print('a e f k m s t')
            print(a, e, f, k, m, s, t)
print('---------------')
for comb in permutations([1, 2, 3, 4, 5, 6, 7, 8, 9, 0], 8):
    a, d, l, m, n, o, s, t = comb
    if all([t, k, s, f]):
        if int(f'{a}{n}{a}{n}{a}{s}') + int(f'{m}{a}{t}{o}{s}{t}') + int(f'{t}{o}{m}{a}{t}') + int(f'{s}{a}{l}{t}') == int(f'{s}{a}{l}{l}{a}{d}'):
            print('a d l m n o s t')
            print(a, d, l, m, n, o, s, t)

Output:


a e f k m s t
4 0 2 5 8 9 6
---------------
a d l m n o s t
1 9 2 3 0 7 5 8

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  • 1
    $\begingroup$ Thats more or less how I came up with them. Are there human-friendly steps to solve them? $\endgroup$ – Per Alexandersson Nov 22 '20 at 7:54
  • $\begingroup$ I have done a sketch of how to do it but writing it all out would be incredibly tedious. $\endgroup$ – Prince Deepthinker Nov 22 '20 at 11:31
  • $\begingroup$ @PerAlexandersson Yep. I'm sticking to my root, though. $\endgroup$ – risky mysteries Nov 22 '20 at 12:48

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