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
 A: For the last equation, $3(100A+10B+C) = 111B$ and hence $100A+C = 27B$. Since $0 \le C \le 9$, you basically need a multiple of $27$ that is very close to a multiple of $100$. Only $27 \times 4$ works, so you know $A,B,C$.
For the second equation you just work from right to left, guessing $M$ and finding $L,K,N$ in order. For example if you guess that $M = 1$, you get that $L = 2$ and $K = 4$ and $N = 8$. But there is more than one solution, actually four in total!
For the first equation, nearly anything goes. We cannot impose that each letter stands for a unique digit, because there are too many letters.
A: The teacher did not see fit to write "each letter stands for a different digit." 
Therefore, "all letters are equal to the number zero" is a valid answer, and should earn maximum points. I would argue that it should earn above maximum, since it also has higher educational value, and homework is all about educational value.
Regarding math, this answer provides an important lesson: the solution to a problem sometimes isn't the most obvious one. Sometimes, when making a proof, a non-obvious approach yields an easier result.
Regarding engineering, this answer provides an even more important lesson: sometimes, the best solution to a problem is to get rid of the problem.
Having to argue why this answer is valid also has educational value.
I think this rather pointless homework provides a wonderful opportunity to teach her several life-saving skills, that will help her immensely in all stages of her life:


*

*Lateral thinking (ie, out of the box).

*Critical thinking (ie, question blind trust in authority)

*And, last but not least, the subtle and holy art of Trolling.


Teaching her at an early age that every word matters will be of immense help when she later has to parse lawyer language in contracts.
Critical thinking means every word written on a piece of paper might not be true. In fact, considering the state of our press, one should never read it without one's bullshit filter set to maximum.
There is also a hidden bonus. When the teacher inevitably grades her paper as FAIL, because most teachers are idiots, your dad will get to file complaint, and have fun in the director's office.
And this will earn him (and you) her unconditional respect.
A: I found by far the easiest way to solve (a) was thru brute force Perl command line:
perl -e "for ($dd=1;$dd<10;$dd++) { for ($ee=0;$ee<10;$ee++) { for ($ff=1;$ff<10;$ff++) { for ($gg=1;$gg<10;$gg++) { for ($hh=0;$hh<10;$hh++) { next if $ee == $ff; next if $dd == $ee; next if $dd == $ff; next if $gg == $hh; $result=qq/$dd$ee$ff/+qq/$ff$ee$ff/; next if $result == 0; if ($result == qq/$gg$hh$hh/) { print qq/$ff $hh $ee $gg $dd\n/; }}}}}} " | sort

Note: this was using ActiveState.com's community version of Perl v5.16.2  (last version that there's a working module for Image::Magick on bribes.org)
Results of the above command line:
1 2 6 4 2
1 2 6 5 3
1 2 6 6 4
1 2 6 7 5
1 2 6 9 7
2 4 7 6 3
2 4 7 7 4
2 4 7 8 5
2 4 7 9 6
3 6 8 5 1
3 6 8 8 4
3 6 8 9 5
4 8 9 6 1
4 8 9 7 2
P.P.S.  Note that the output is in F H E G D order as I was comparing results to the answer above mine =)
A: 1.This is a homework for a third grade girl, so we need to think as a kid but not as an advanced expert, in this case we will be able to explain to this kid step by step how we solve the problem and he will understand the solution in an easy way.
2.I do not think that the purpose of the teacher of a third grade was to make all letters equal to 0 because it will not make any sense to the children and they will not learn anything from this case, so I think that this is not a valid solution at all.
3.FYI: the solutions must be from right to left because this is how children learn to do calculations.
4.The three operations seems to be independents because they are no common letters between them.
5.I have started to solve the third operation (c): 

The trick is to firstly we give the value 8 to C then we can find B because 8+8+8=24 which means we write 4 and we put 2 on the top of the second column. Therefore B = 4 then we replace B by 4 in the second column and in the result. In the second column we get 2+4+4+4=14 we write 4 and we put 1 on the top of the last column, in the last column we have the result 4 and we already have 1 on the top so A will be 1 (1+1+1+1=4)
SO A=1 ,B=4 ,C=8
6.For the second operation (b):

We give the value 1 to M then L= M+M = 1+1 = 2 so K = L+L = 2+2 = 4 in the end N = K+K = 4+4 = 8 therefore M=1, L=2, K=4, N=8 .Please notice if we give to M a value of 2,4,6,7,9 so the result will be on more then 3 digits (e.g. M=2 the result will be 1684) Therefore the solutions for M are 1,3,5,8.
A: 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.
A: There is no unique answer to (a).  The intention must be to expose grade school students to normal adult frustrations. To find all those answers start with F. Other answers have shown that it must be 1,2,3 or 4.  For each F there is a unique H (2,4,6,8) and E (6,7,8,9), respectively. Again, the other answers get that far. Then G can be any number that isn't F or H or E. There is a unique D for each G, but some of them are 0 (criptarithms are understood to not have leading 0s) and others are negative.
The solutions are:
F   H   E   G   D
1   2   6   5   3
1   2   6   7   5
1   2   6   9   7
2   4   7   6   3
2   4   7   8   5
2   4   7   9   6
3       6       8       5       1
3       6       8       9       5
4   8   9   6   1
4   8   9   7   2
I used a spreadsheet.
I thought the answer to (b) was easy to get by inspection. 
Several answers have the same answer as me to (c): A,B,C are 1,4,8.
A: The homework is badly written.  It should tell you that each letter represents a different number from 0-9 and you are supposed to figure out what numbers the letters mean.  It's supposed to be a math problem about thinking about what 'carrying the one' means when two numbers sum to larger than 10.  Because the problem is poorly worded your child is not well-equipped to solve the problem on her own (there is no particular reason from the perspective of a 8-year-old child why that is a good rule and not, e.g. the alphanumeric rule you postulated or some variety of word game e.g. DEF +FEF = GHH, GHH is the sound one makes when frustrated, so the problem is about how you can't add words, only numbers).  
When my younger sister struggles with such a problem, I review the question to ensure there's no instructions she missed that would make the problem clear, then ask her to write a short explanation as to why the problem doesn't make sense instead of answering it (which I hope will prepare her for higher math where proving that the question asked is impossible is an expected approach for certain kinds of problems).  She gets practice thinking about math and how to interpret different kinds of questions and tell what part of something she's doing is the 'hard' part, which I think more than makes up for whatever she's 'supposed' to be getting out of such poorly worded questions.
My sister is now in 8th grade, and asking her to write down what makes things hard in 3rd grade would probably not have gone as well.  Writing is hard for a 3rd grader!  When she was younger, we would just talk about what made the problem hard for her and I would try and get her just to get past 'I/you can't do it/don't know how to do it' (the immediate answer of any 3rd grader told to do something hard) and onto the reasons why she felt she couldn't do it.  I'd then either show her what I thought the teacher wanted her to do or just have her leave it blank and tell her it's ok not to be able to do the problems as long as you try.  
Be aware that figuring out why something is hard is itself really hard sometimes, especially for children, and especially the first several times they are asked to do it.  You'll probably spend at least 30 minutes on that question and you'll need to spend a significant part being quiet-but-supportive/understanding while your child thinks.
A: Each letter stands for a digit, but not all of them are different. For example, if $A=1, B=4$ and $C=8$, that makes the last equation correct. It's more than standard third grade math, it's a puzzle to try to figure out which letters are which digits.
A: It is clear that problems a), b), c) are independent each other.
I find just solutions maybe not in an exhaustive way.
a) DEF+FEF=GHH
$DEF+FEF=GHH\implies F+F=H \text { or }1H$
$F+F=H\implies (F,H)\in\{(0,0),(1,2),(2,4),(3,6),(4,8)\}$
$F+F=1H\implies(F,H)\in\{(5,0),(6,2),(7,4),(8,6),(9,8)\}$
We discard the case $(F,H)=(0,0)$ because it gives the "solutions" $100+000=100$ and $150+050=200$ and we don't consider $000$ nor $050$.
Now $(F,H)=(1,2)$ gives $DE1+1E1=G22\implies E=1$ and $D+1=G$ which gives the solutions $(D,G)\in\{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9)\}$.
We have in this case the sum $D11+111=G22$ with the consequent eight given values for $(D,G)$.
b) KLM+KLM=NKL
$2*(KLM)=NKL\implies L\in\{0,2,4,6,8\}$
$L=0\implies M=5\text{ or } 0\implies 2*(K05)=N10\implies K=1\implies N=2$.
(the value $M=0$ gives discarded trivial solution).
We have the sum $105+105=210$ as a solution.
c) ABC+ABC+ABC=BBB
We have $3*(ABC)=111*B\implies ABC=37*B\implies B\gt 2\implies B=4$ because for $B\in\{3,5,6,7,8,9\}$ we would have $37*B=XYZ$ with $Y\ne B$.
Now $3*A4C=444\implies A4C=148\implies (A,C)=(1,8)$
In this case we have $148+148+148=444$.
A: 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!!!
A: A good place to start would be with $ABC + ABC + ABC = BBB$.  Since $BBB$ can only take one of $6$ values (it must be at least $123+123+123 = 369$ so it's one of $444,555,\ldots,999$), $ABC$ is uniquely determined as one-third of $BBB$.  The fact that $B$ appears in both numbers narrows down the field even further, as shown in Billel Hacaine's answer: $A=1, B=4, C=8$.
From $DEF + FEF = GHH$ we have from the one's digit that $H$ is even, hence there is no carry and thus $2F = H$.  Since the last digit of $2E$ is also $H$ but $F \ne E$, it follows that $E = F+5$.  There us no choice of $F$ for which $\{F,F+5,2F\}$ is disjoint from $\{1,4,8\}$, though.
A: There are different ways to solve this. 
Given the age and school year, I think a logical trial and error approach is what they expect. 
Assume that any number is possible for any character, then try them to see what follows. Knowing basic addition, the possibilities are far fewer than it may seem at first.
I'm starting with the ABC problem, since it looks to me like the easiest one (many repetitions, only 3 letters).
Since addition starts with the smallest value (C) that's where we begin testing numbers.
If C=0 then since C+C+C=B we get B=0, giving
 A00
+A00
+A00
----
 000

The only possibility here is A=0, and this is a valid solution but too simple so I'd say we can ignore it and look for another.
If C=1 then B=3 and we get
 A31
+A31
+A31
----
 333

The second column isn't correct since 3+3+3=9 so it's safe to say C is not 1.
Continue like this.
C=2 => B=6 same problem
C=3 => B=9 same problem
C=4 => B=2 (now we have a 1 carry) which gives for column 2: 1+2+2+2=2 , also wrong
You'll find that the only possibility is C=8 which gives you
 12
 A48
+A48
+A48
----
 444

now you solve for A which is 1 since 1+A+A+A=4
For the KLM problem try 1 for M giving
 421
+421
----
 842

Or M=3 giving
  1
  263
 +263
 ----
  526

Or M=5 giving
   1
  105
 +105
 ----
  210

There are a few possibilities here (I didnt try all of them). If M = 2 or 4 on the other hand, you end up with a carry in the first column giving you a 4 digit sum, which isn't the case, so these are not options. 
For the DEF problem, assuming that different letters represent different numbers:
One solution is
 1     
 361
+161
----
 522

A: Since there seems to be no answer to the second equation yet I will have a go at it:

For this is a third grade task I will be assuming that all variables have positive natural numbers as values.

We can see that $N$ is a single digit. Therefore we know that $N < 10$. We also know that $2K = N$. If we substitute the $N$ in the first equation for $2K$ and devide the whole formula by $2$ we get $K < 5$.

The other two digits can now be determined similarly: $K < 5$ and $K = 2L$, therefore $L < 2.5$. Since we are working with natural numbers this is equivalent to $L \le 2$.

$L \le 2$ and $L = 2M$, therefore $M \le 1$.

So $M$ can either be $0$ or $1$. However, if $M$ was $0$, the equation $2M = L$ would have to be $2M = M$ instead (I am quity sure that a third grade exercise won't have two Variables with the same value). That leaves us with only one conclusion: $M = 1$. With that knowledge calculating the other variables is easy:

$M=1$

$L=2M = 2$

$K = 2L = 4$

$N = 2K = 8$
