Solving the cryptarithm $\mathrm{TWO}+\mathrm{TWO}=\mathrm{FOUR}$ in base $10$ and base $7$

Question

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.

Answer 1

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.

Answer 2

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:

$$\begin{array}{ccccccc} &&&&5&2&3\\ +&&&&5&2&3\\ \hline &&&1&3&4&6\\ \end{array}$$