TITLE: Solving $DEF+FEF=GHH$, $KLM+KLM=NKL$, $ABC+ABC+ABC=BBB$
QUESTION [73 upvotes]: 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

REPLY [2 votes]: I am assuming that we have three separate problems, we are in number base 10, that there are no leading zeroes, and that there is a one-to-one correspondence between digits and letters.
For the first problem, observe that H is even because F+F=H.  Also note that F+F does not generate a carry, because E+E is also even.   So there are four possibilities for (F,H,E): (1,2,6),(2,4,7),(3,6,8),(4,8,9).  Each of these has several possibilities for D and G, leasing to 13 solutions for (F,H,E,D,G):
(1,2,6,3,5), (1,2,6,5,7), (1,2,6,7,9); (2,4,7,0,3), (2,4,7,3,6), (2,4,7,5,8), (2,4,7,6,9); (3,6,8,0,4), (3,6,8,1,5), (3,6,8,4,7); (4,8,9,0,5), (4,8,9,1,6), (4,8,9,2,7).  If we assume that the solution is unique, other number bases, starting with 5, could be explored.
For second problem, there are four possibilities for K because K+K does not generate a carry. Also, L must be even. An even guess for K rules out a carry for M+M, while an odd guess for K requires a carry for M+M. This greatly limits the possibilities to consider. There are four possible values for K, and the solutions (K,L,M,N) are (1,0,5,2), (2,6,3,5), (3,6,8,7), (4,2,1,8), and (4,6,3,9).
Four the third problem, note that there are nine possible numbers of the form BBB.  But one third of BBB is a three digit number only for B>2.  If you divide each of the three digit numbers 333, 444, 555, etc. by 3 only one of them, 444, gives a quotient with the same middle digit.   So (A,B,C)=(1,4,8).
A great book that might help with this is beyond a third grade student, but might help us seniors with Altzheimer's: Doerfler's Dead Reckoning.
Another method of solution would be a generate and test procedure on a programmable calculator.  My favorite would be the WP-34s.  Although it runs on a discontinued platform, you can emulate it on your iPhone.   
What?  A third grade girl with no iPhone?   That is child abuse!!!