## Task 21/82

We are looking for all four-digit numbers (in decimal notation) with the digit sequence $\{a ; b ; c ; d\}$ where $a ; b ; c ; d \in N, 1 \leq a \leq 9,0 \leq b ; c ; d \leq 9$, whose sixfold has the digit sequence $\{a ; a ; c ; b ; d\}$.
According to the problem statement,

$$
6(1000 a+100 b+10 c+d)=11000 a+100 c+10 b+d
$$

or (after appropriate rearrangement): $5 d=10(500 a+4 c-59 b)$.

It immediately follows that $d$ is an even number: $d=2 e$ with $e \in N, 0 \leq e \leq 4$; thus, the equation takes the form

$$
e=500 a+4 c-59 b \quad \text { or } \quad 59 b=500 a+4 c-e
$$

Since $a \geq 1$, it follows that $59 b \geq 496$, so $b>8$. Therefore, $b=9,531=500 a+4 c-e$.

Since $4 c-e<100$, it follows that $a=1$ and from this

$$
31=4 c-e \quad ; \quad c=7+\frac{3+e}{4}
$$

Thus, $e=1, d=2, c=8$. Indeed, $6 \cdot 1982=11892$.