## Task 24/89

Let $n \in N$ be a truly four-digit number in the decimal system; its representation in the positional system with base $b \neq 10 \quad(b \in N, b \geq 2)$ is 1549.

This representation differs from the one in the decimal system exactly at the two middle positions. Calculate $b$ and $n$.
Let $a_{1} ; a_{2} \in N, 0 \leq a_{1} ; a_{2} \leq 9$ be the unknown digits of the two middle positions in the decimal system and $b=10+k$ with an integer $k \geq 1$ (since the digit 9 appears in the representation in the $b$-system, $b>9$ and because $b \neq 10$, $b \geq 11$). Then the equation

$$
n=1000+100 a_{2}+10 a_{1}+9=(10+k)^{3}+5(10+k)^{2}+4(10+k)+9
$$

Simplified and divided by 10, we get

$$
10 a_{2}+a_{1}=3 k^{2}+40 k+54+\frac{k^{3}+5 k^{2}+4}{10}
$$

For the left side of this equation to remain a two-digit number, $k<2$, so $k=1$. Thus, $10 a_{2}+a_{1}=$ $98 ; b=11$ and $n=1989$.

### 2.30 Problems and Solutions 1990