4. Find all four-digit numbers that, when divided by 197, give a remainder of 47, and when divided by 198, give a remainder of 37.
4. Let $n$ be the desired number. Then we have $n=$

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Ok 2017 1B 3 $197 k+47=198 l+37$ for some non-negative integers $k$ and $l$. From this, it follows that $198 l-197 k=10$. Since $198-197=1$, one obvious solution is $k=l=10$, and then we have $n=2017$. Suppose there is another solution. Then, if we subtract the equation $198 \cdot 10-197 \cdot 10=10$ from the equation $198 l-197 k=10$, we get $198(l-10)-197(k-10)=0$. Since 197 and 198 are coprime, it follows that $197 \mid l-10$, i.e., $l=197 s+10$ for some $s$. If $s>0$, we would have $l \geqslant 207$ and $n=198 l+37 \geqslant 198 \cdot 207+37 > 9999$, i.e., $n$ would not be a four-digit number. Therefore, the only solution is $n=2017$.