4.46. 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.
4.46. Answer: 1946. Let $N$ be the desired number. By the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. Moreover, $N<10000$, so $l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75$. Further, $131 k+112=132 l+98$, therefore $131(k-l)=l-14$. Consequently, if $k \neq l$, then $|l-14| \geqslant 131$. But $l \leqslant 75$, so $k=l$ and $l-14=0$. Thus, $N=131 \cdot 14+112=132 \cdot 14+98=1946$.