A student marked the product of an integer and 467 in their homework as 1925 817. The teacher underlined the 9 and the 7 as errors. What are the correct digits and what was the multiplicand?
I. Solution: The faulty result, when divided by 467, leaves a remainder of 376; the two faulty digits must be changed so that this disappears. The value at the 1's place can decrease by at most 7, or increase by 2, which means the main change must be achieved at the $10^{5}$ place. Here, only a reduction can correct the issue. Since the remainder of the division $10^{5} : 467$ is 62, each unit decrease in the faulty 9 results in a decrease of the remainder by 62. Such a step is needed as many times as the quotient of 376 : 62, which is approximately 6, so only $9-6=3$ can replace the 9. The remaining $376-6 \cdot 62=4$ units of reduction can be achieved by changing the 1's place digit from 7 to $7-4=3$. Therefore, the correct result is 1325 813, and the multiplicand is one 467th of this: 2839.

Kirschner Miklós (Bp. II., Rákóczi g. IV. o. t.)

Note. The unique solution was made possible, on the one hand, by the fact that the 62-unit "steps" are much longer than the 7-unit change possible at the 1's place; on the other hand, by the fact that we would only reach a number that, when divided by 467, the quotient is close enough to an integer to be corrected at the 1's place, after another 15 steps. Indeed, 467 itself is "far" from the 7- or 8-fold of 62, which encloses it: 33 or 29 units away, while its double, 934, is "close" to $15 \cdot 62=930$.

II. Solution: Let the correct digit in the place of the faulty 9 and 7 in the correct result be $a$ and $b$, respectively, and the multiplicand be $c$, where $0 \leqq a \leqq 8$ (since the 9 is faulty), and $0 \leqq b \leqq 9, b \neq 7$. Then

$$
1025810+10^{5} a+b=467 c
$$

and thus

$$
c=\frac{1025810+10^{5} a+b}{467}=2196+214 a+\frac{62 a+b+278}{467}
$$

The smallest and largest values of the fractional expression in the latter form are 278/467, which is greater than 0, and 783/467, which is less than 2, when $a=b=0$ and $a=8, b=9$, respectively. However, $c$ is an integer, so the only possible value of this fractional expression is the only integer between these limits: 1. Now, from $62 a+b+278=467$, we get $b=189-62 a$, so $0 \leqq 189-62 a \leqq 9$, hence

$$
\frac{180}{62} \leqq a \leqq \frac{189}{62}
$$

thus $a=3$, from which $b=3$, and finally $c=2839$.