Solving the cryptarithm $\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$ in base $10$ and base $7$ 
Find the numerical value of each of the letters in the following expression $$\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$$
in (a) base $10$ and (b) base $7$.

I don't even know how I would approach this problem. Any help would be appreciated.
 A: I do not know what 102152111's "brute force" solution is for base 7, but we can use a relatively small number of trials by focusing on $O$ as our first target.  Since $F=1$, the distinctness criterion rules out that value for $O$, so we try the six others in turn.  Each such value of $O$ corresponds to a unique value for $T$ and a unique carry digit from the $7^1$ column to the $7^2$ column, thus simplifying the analysis.

*

*$O=0$ fails quickly, since it would give $R=0$ in the units place.


*$O=2$ corresponds to $T=4$ and a carry digit of $1$ ($4×4+1=12$ base $7$), but then it leads to $R=4$ in the units place which fails because $T=4$.


*$O=3$ corresponds to $T=5$ and a carry digit of $0$, and leads to $R=6$ in the units place.  We are good so far on distinctness.  To get the right carry digit into the $7^2$ place for this case we then need $W<4$.  $W=1$ and $W=3$ fail distinctness from previously determined letters, $W=0$ gives $U=0$ because $2O<7$, but $W=2$ makes it through.  This is the solution $523+523=1346$ given by 102152111.


*$O=4$ gives $R=1$ in the units place, which gets an $F$, so to speak.


*$O=5$ corresponds to $T=6$, a carry digit of $0$ into the $7^2$ place, and $R=3$.  This looks good so far, but  no values of $W$ work.  $W=3$ is too large because $35+35>100$ base $7$; $W=1$ is not distinct from $F$; $W=0$ and $W=2$ fail to hold up distinctness when we calculate $U$.  So $O=5$ falls just short of making it through.


*$O=6$ requires $T=6$ and a carry digit of $1$, the former of which already fails distinctness.
So $523+523=1346$ derived from the $O=3$ case is the unique solution.
A: For (a), base 10:
The easiest place to start with is F (in four) and that must be 1, as

TWO + TWO

when the letters are at the highest possible value (987+987) = 1974, so

F=1

Now we move to O.
$O$ cannot be 1,
so can be $2,3,4,5,6,7,8,9,0$
If $O$=2, then $R$ is 4. $W$ cannot be assigned a number, as if $W$ is 5 and more, we will have a carry over into $T$ and 11 (12-1) cannot be divisible by 2, so then no number will be assigned to $T$, so 5 or more cannot be correct.
$W$ cannot be 1 because of $F$. $W$ cannot be 2 because of $O$. If $W$ was 3, then $U$=6 but $T$=6 (12 divided by 2), so it cannot be 3. $W$ cannot be 4 because $R$ is 4.
So, $O$ isn't 2.
If $O$ was 3, then $R$=6. If $R$= 6 then $T$ cannot be assigned a number, since if there is a carry over of 1, 12 divided by 2 is 6, but $R$ is 6.
SO, $O$ isn't 3.
If $O$ is 4, then $R$=8.
Now we have
$$\begin{array}{ccccccc}
&&&&T&W&O\\
+&&&&T&W&O\\
\hline
&&&F&O&U&R\\
\end{array}$$
=
$$\begin{array}{ccccccc}
&&&&?&?&4\\
+&&&&?&?&4\\
\hline
&&&1&4&?&8\\
\end{array}$$
If there was a carry over into the $T$$T$$O$ column, then 13 divided by 2 is not a whole number. So $T$ has to be 7.
Now,
$$\begin{array}{ccccccc}
&&&&7&?&4\\
+&&&&7&?&4\\
\hline
&&&1&4&?&8\\
\end{array}$$
Now, we find $W$ and $U$.
$W$ cannot be 0, 1 and 4 from just looking, so we have $2,3,5,6,7,8,9$
If $W$ was 2, then $U$= 4 but $O$ is 4.
If $W$ was 3, then $U$= 6 and that is the answer.
$$\begin{array}{ccccccc}
&&&&7&3&4\\
+&&&&7&3&4\\
\hline
&&&1&4&6&8\\
\end{array}$$
This is one solution. Other solutions are:
765+765==1530
836+836==1672
846+846==1692
867+867==1734
928+928==1856
938+938==1876
For b)
I used brute force for this, and found an unique solution:
$$\begin{array}{ccccccc}
&&&&5&2&3\\
+&&&&5&2&3\\
\hline
&&&1&3&4&6\\
\end{array}$$
