(1) In an arithmetic sequence $\left\{a_{n}\right\}$ with a common difference of $d$ and all terms being positive integers, if $a_{1}=1949$, $a_{n}=2009$, then the minimum value of $n+d$ is $\qquad$ .
(1) 17 Hint: From $2009=1949+(n-1) d$, we get $(n-1) d=60$. When $n \geqslant 3$, then the possible values of $\{n-1, d\}$ are $\{2,30\},\{3,20\},\{4$, $15\},\{5,12\},\{6,10\}$, so $d+n-1 \geqslant 16$, which means $n+d \geqslant 17$.