The original problem: When a natural number is multiplied by the number one greater than itself, the product is in the form $A B C D$, where $A, B, C, D$ are different digits. Starting from the number three less than the original, the product is in the form $C A B D$. Starting from the number thirty less than the original, the product is in the form $B A C D$. Determine these numbers.

The corrected problem, published in September 1966 (with a two-letter swap at the end) - we will publish the solution in the March 196\% issue:

When a natural number is multiplied by the number one greater than itself, the product is in the form $A B C D$, where $A, B, C, D$ are different digits. Starting from the number three less than the original, the product is in the form $C A B D$. Starting from the number thirty less than the original, the product is in the form $B C A D$. Determine these numbers.
Denoting the smaller factor of the first product by $x$, the requirements are:

$$
\begin{aligned}
& x(x+1)=x^{2}+x \quad=1000 A+100 B+10 C+D, \\
& (x-3)(x-2)=x^{2}-5 x+6=1000 C+100 A+10 B+D, \\
& (x-30)(x-29)=x^{2}-59 x+870=1000 B+100 C+10 A+D .
\end{aligned}
$$

Subtracting (2) from (1):

$$
\begin{aligned}
6(x-1) & =900 A+90 B-990 C=90(10 A+B-11 C) \\
x-1 & =15(10 A+B-11 C) \\
x & =15 m+1
\end{aligned}
$$

where $m$ is an integer.

All three of our products are four-digit numbers, and the leading digits $A, B, C$ are not 0. Therefore, from (1) $x1000
$$

It is more convenient to calculate if we also increase the first factor to $x-29$, because then the left side increases:

$$
(x-29)^{2}>1000, \quad \text { so } \quad x-29 \geqq 32, \quad x \geqq 61
$$

because the smallest four-digit square number is $1024=32^{2}$.

The lower limit obtained for $x$ is precisely a number of the form (4). However, $x$ cannot be 61, because in (3) $31 \cdot 32<1000$, the increase we made just crossed the smallest four-digit number. Thus, $61<x<100$, and in (4) only the values $m=5$ and $m=6$ are possible.

$m=5$ does not work, because when $x=76$, $76 \cdot 77=5852$ has two identical digits.

$m=6$ provides a solution:

$$
91 \cdot 92=8372, \quad 88 \cdot 89=7832, \quad 61 \cdot 62=3782
$$

from all of which $A=8, B=3, C=7, D=2$. There is no other solution to the problem.