Can the values of the letters as digits be chosen in the following two (independent) problems so that the statements are true?
a) $\mathrm{FORTY}+\mathrm{TEN}+\mathrm{TEN}=\mathrm{SIXTY}$

b) DREI+DREI+DREI=NEUN

(Identical letters are to be replaced by the same digits, different letters by different digits. The a) problem is in English $40+10+10=60$, the b) problem is in German $3+3+3=9$.)
a) Rewriting the addition in the usual columnar form, from the last two columns

$$
\overline{T Y}+2 \cdot \overline{E N}=\overline{T Y}+100 s, \quad \overline{E N}=50 s
$$

where $s$ is the carry into the hundreds column; its value cannot be less or more than 1, since $E \neq N$, and thus $098$; so $4 \leq R \leq 5$.

The value $R=4$ excludes the middle triplet (9), and

$$
7 E+U=22
$$

from which either $E=2$ and $U=8$ or $E=3$ and $U=1$. The first pair leaves only the first triplet (9):

$$
\begin{aligned}
D=0, \quad I=7, \quad \overline{D R E I} & =0427, \text { and thus } \\
N E U N & =1281
\end{aligned}
$$

which, considering the leading zero, can only be accepted as a broader sense solution.

The pair $E=3$ and $U=1$ leaves only the last triplet (9), which is a real solution:

$$
2439 \cdot 3=7317
$$

For the value $R=5$, from (10) $7 E+U=52$, which is only satisfied by the pair $U=3, E=7$, and with the middle triplet (9) we again get a real solution:

$1578 \cdot 3=4734$

$\beta$ ) Similarly, for $s=2$, $e=1$, from (4) $4 \leq I \leq 6$, from (8) $I=D+4$, and for $D, I, N$ two triplets of digits are possible:

$$
0,4,22,6,8
$$

(since the value $I=5$ leads to $I=N$). The equivalent of (10) is:

$$
30 R=7 E+U+199>199, \quad R \geq 7
$$

From the attempt with $R=7$, $7 E+U=11$, which is only satisfied by $E=1, U=4$ and the second triplet (11) gives another real solution:

$2716 \cdot 3=8148$.

Similarly, from $R=8$, $7 E+U=41, E=5, U=6$, and it is not a real solution:

$$
0854 \cdot 3=2562
$$

Finally, for $R=9$, $7 E+U=71$, which is only satisfied by $U=8, E=9=R$.

c) According to all this, we have found three real and two broader sense solutions for the German problem. $\left(\mathrm{V}_{1}\right)-\left(\mathrm{V}_{3}\right)$, and $\left(\mathrm{T}_{1}\right),\left(\mathrm{T}_{2}\right)$, respectively, and only one solution for the English problem.

Remarks. 1. In part a), we can also proceed as follows, knowing the (1) relationship. We cannot choose the members of the $F, S$ pair from the numbers 6, 7, 8, because then only one of these would remain for the roles of $R$ and $T$, and the largest of them, 8, with the next 4-digit number, would only give 20 for the $(R+T+T)$ sum, from which $s=1$ would imply $X=1$, but 1 is already occupied.

Therefore, the members of the $F, S$ pair are from the digits 2, 3, 4: either $F=2, S=3$, or $F=3, S=4$. The sum $(T+T+R)$ is at most $8+8+7=23$, so with $s=1$, $X$ is at most 4, and from the two cases above, in the first case $X=4$, in the second case $X=2$. If $X=4, R+T+T=23$, so $R=7, T=8$, and $Y$ can only be the unused 6:

$$
29786+850+850=31486
$$

If $X=2$, so $R+T+T=21$, which is not possible ( $T=8$ is occupied by $R=5$, $T=7$ leads to $T$ equal to $R=7$). 2. In part b) of the problem, the acceptance of zero as a leading digit does not raise such further branching questions as in the case of the recent 1378. exercise 1[^0]


[^0]:    ${ }^{1}$ K. M. L. $44(1972) 119$. o.