Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98.

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Let $N$ be the desired number. According to the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. From this, $131(k-l)=l-14$. Denoting $k-l=t$, we have $l=131 t+14, N=132 \cdot 131 t+132 \cdot 14+98$. Since the number $132 \cdot 131$ is a five-digit number, only $t=0$ fits, giving $N=132 \cdot 14+98=1946$.

## Answer

1946.