35. What is the smallest positive integer $n$, where $n \neq 11$, such that the highest common factor of $n-11$ and $3 n+20$ is greater than 1 ?
35. Answer: 64
Let $d>1$ be the highest common factor of $n-11$ and $3 n+20$. Then $d \mid(n-11)$ and $d \mid(3 n+20)$. Thus $d \mid\{3 n+20-3(n-11)\}$, i.e., $d \mid 53$. Since 53 is a prime and $d>1$, it follows that $d=53$. Therefore $n-11=53 k$, where $k$ is a positive integer, so $n=53 k+11$. Note that for any $k, 3 n+20$ is a multiple of 53 since $3 n+20=3(53 k+11)+20=53(3 k+1)$. Hence $n=64$ (when $k=1$ ) is the smallest positive integer such that $\operatorname{HCF}(n-11,3 n+20)>$ 1.