When a three-digit number is multiplied by 7, the product has 638 in the thousands place. Which of these three-digit numbers?
I. Solution. The number $N$ we are looking for has 4 in the units place; this is the only number that, when multiplied by 7, results in 8 in the units place.

In $7N$, the number in the tens place is 3: this includes the two tens from $7 \times 4 = 28$. If we subtract this: $3 - 2 = 1$; therefore, the number in the tens place of $N$ can only be 3.

In $7N$, the number in the hundreds place is 6; this includes the two hundreds from $7 \times 3 = 21$. Subtracting this from 6: $6 - 2 = 4$; thus, the number in the hundreds place of $N$ can only be 2.

Accordingly,

$$
N = 234
$$

Csató Aliz (Ev. leányg. IV. o. Kószeg.)

II. Solution. The number we are looking for, when multiplied by 7, is: $1000x + 638$, so

$$
N = \frac{1000x + 638}{7} = 142x + 91 + \frac{6x + 1}{7}
$$

This will be an integer if

$$
6x + 1 = 7y, \quad x = \frac{7y - 1}{6} = y \quad (1, 2, 3, \ldots)
$$

But

$$
x = \frac{7y - 1}{6} = y + \frac{y - 1}{6}
$$

is only an integer if

$$
y - 1 = 6z, \quad y = 6z + 1 \quad (z = 0, 1, 2, 3, \ldots)
$$

Thus,

$$
x = 7z + 1
$$

and

$$
\begin{gathered}
N = 142(7z + 1) + 91 + \frac{6(7z + 1) + 1}{7} = 1000z + 234 \\
(z = 0, 1, 2, 3, \ldots)
\end{gathered}
$$

However, $N$ is a three-digit number, so $z = 0, x = 1, 7N = 1638$ and

$$
N = 234
$$

Fehér Ödön (Érseki g. V. o. Bp. II.)