<s>[INST] Solving $DEF+FEF=GHH$, $KLM+KLM=NKL$, $ABC+ABC+ABC=BBB$ She visits third class and is $8$ years old (you can imagine how ashamed I felt when I said so to her). I helped her with lots of maths stuff today already but this is very unknowable for me. Sorry it's in German but I have translated it :)

It's saying "Each letter represents a digit. Determine them". First question, what is "them"? The letters I guess?
How shall I determine them when they are unknown? Or is it simply $A=1, B=2, C=3, D=4, E=5, F=6, G=7, H=8, K=11, L=12, M=13, N=14$?
Alright...
With this we gave a) the first try:

It doesn't seem to make sense to set $A=1, B=2, ...$
Or we did something wrong.. Any ideas how this could be solved? :s
 [/INST] For the 3rd sum:Let the carry from the 1st column to the 2nd column be $d$, and let the carry from the 2nd column to the 3rd column be $e$.
The second column implies that $3B+d $ is equal to $B$ plus a multiple of $10,$ so $2B+d$ is a multiple of $10$. So $d$ is even, so $d=0$ or $d=2$.
If $d=0$ then $2B=2B+d$ is a multiple of $10$ so $B=0$ or $B=5$.  Now $B=0$ gives the "trivial" solution $A=B=C=0$ (because $3\cdot ABC=BBB=000=0$.) But if $d=0$ and $B=5$ then $3C=B+10d=B=5$, which is impossible.
So a no-trivial solution must have $d=2.$ The $2B+2=2B+d$ is a multiple of $10$, which requires $B=4$ or $B=9$. 
But if $d=2$ and $B=9$ then  $3C=10d+B=20+9=29,$ which is impossible.
With $d=2$ and $B=4$ we have $3C=10d+B=20+4=24$, giving $C=8.$ Finally $3B+d=12+2=14$, giving the carry $e=1.$ And $3A+1=3A+e=B=4,$ giving $A=1$. (Solution:$148+148+148=444$). 
8-year-olds who can solve this on their own would be rare.
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