Problem 10.1. (a) (2 points) A natural number $n$ is less than 120. What is the largest remainder that the number 209 can give when divided by $n$?

(b) (2 points) A natural number $n$ is less than 90. What is the largest remainder that the number 209 can give when divided by $n$?
Answer: (a) 104. (b) 69.

Solution. Let $209=n k+r$, where $k-$ is the quotient, and $r$ is the remainder of the division. Since $rn k+r=209=n k+r>r k+r=r(k+1)$, hence

$$
k+1>\frac{209}{n} \quad \text { and } \quad r\frac{209}{119}$, i.e., $k \geqslant 1$. Then $r\frac{209}{89}$, i.e., $k \geqslant 2$. Then $r<\frac{209}{3}$, i.e., $r \leqslant 69$. The value $r=69$ is possible when $n=70$ and $k=2$.