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In the following addition setup, each letter correspond to a unique digit. Find the different values of the letters. Requirement: r is not equal to 0.

   c, a, k, e
   b, i, t, e
      t, e, a
+     b, u, n
--------------
b, r, e, a, k

The solution is unique (provided r is non-zero).

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  • $\begingroup$ It is then not a "hint", but a requirement. $\endgroup$ – WhatsUp Oct 25 '20 at 20:18
  • $\begingroup$ @WhatsUp yes, that's more accurate, so I changed that word for now. $\endgroup$ – Per Alexandersson Oct 25 '20 at 21:24
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Firstly,

$b = 1$.

Secondly,

the sum is smaller than $10000 + 1900 + 800 + 200 < 13000$, hence $r = 2$ and $c = 9$.

Now

write down the equality as $$110 t + 100 i + 91 a + 10 u + 9 k + n - 88 e = 1900.$$ This tells us that $t > 3$, otherwise the left hand side is at most $1876$.
The same argument shows that $e \leq 3$.

We then

look at the above identity mod $9$. It becomes $$2t + i + a + u + n + e \equiv 1 \mod 9.$$ But we also know that $$t + k + i + a + u + n + e = 0 + 3 + 4 + 5 + 6 + 7 + 8 \equiv 6 \mod 9,$$ therefore $t - k \equiv 4\mod 9$, i.e. either $t - k = 4$ or $k - t = 5$.
However, we have seen that $t > 3$, hence $t - k = 4$.

After this, we look at

the possible values of $e$. If $e = 3$, then we must have $t = 8$ and $k = 4$ (the other possibility, $t = 4$ and $k = 0$, will result a value that is too small) and we see that no possible choices of $a$ and $n$ can make the last digit to work.
Therefore $e = 0$ and $(t, k) = (8, 4)$ or $(7, 3)$.

It is clear that

$(t, k) = (8, 4)$ doesn't work, as the last digit causes a problem.
Thus $t = 7$ and $k = 3$.

The remaining is clear. Final answer:

$e, b, r, k, u, a, i, t, n, c = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

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Here is how you can use python brute forcing to find the answer:

from itertools import permutations

for r in range(1, 10):
    for comb in permutations([1, 2, 3, 4, 5, 6, 7, 8, 9, 0], 9):
        a, b, c, e, i, k, n, t, u = comb
        if r not in [a, b, c, e, i, k, n, t, u]:
            if int(f'{c}{a}{k}{e}') + int(f'{b}{i}{t}{e}') + int(f'{t}{e}{a}') + int(f'{b}{u}{n}') == int(f'{b}{r}{e}{a}{k}'):
                if all([i != 0 for i in [c, b, t]]):
                    print(a, b, c, e, i, k, n, r, t, u)

Output:

5 1 9 0 6 3 8 2 7 4

So just like the other answers, it's

a, b, c, e, i, k, n, r, t, u = 5 1 9 0 6 3 8 2 7 4

Technically, numbers like 0120 are still numbers, so if leading zeros are allowed, the code would be:

from itertools import permutations

for r in range(1, 10):
    for comb in permutations([1, 2, 3, 4, 5, 6, 7, 8, 9, 0], 9):
        a, b, c, e, i, k, n, t, u = comb
        if r not in [a, b, c, e, i, k, n, t, u]:
            if int(f'{c}{a}{k}{e}') + int(f'{b}{i}{t}{e}') + int(f'{t}{e}{a}') + int(f'{b}{u}{n}') == int(f'{b}{r}{e}{a}{k}'):
                 print(a, b, c, e, i, k, n, r, t, u)

Output:

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

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