0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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)

| improve this answer | |
New contributor
tsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
0
$\begingroup$

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 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}'):
        if all([i != 0 for i in [t, k, s, f]]):
            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 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}'):
        if all([i != 0 for i in [t, k, s, f]]):
            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

| improve this answer | |
$\endgroup$
  • 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 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 at 11:31
  • $\begingroup$ @PerAlexandersson Yep. I'm sticking to my root, though. $\endgroup$ – ention everyone Nov 22 at 12:48
0
$\begingroup$

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

101015

+ 318758

+ 87318

+ 5128

= 512219

| improve this answer | |
$\endgroup$
0
$\begingroup$

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).

| improve this answer | |
New contributor
Vipul Tawde is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • 1
    $\begingroup$ "A and T are either both odd or both even" -> Why not just A+T=10. $\endgroup$ – tsh Nov 22 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 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.