19. Let $a_{1}, a_{2}, \ldots, a_{11}$ be positive integers with $a_{1}<a_{2}<\cdots<a_{11}$. If $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \ldots, \frac{1}{a_{11}}$ form an arithmetic sequence (i.e. $\frac{1}{a_{2}}-\frac{1}{a_{1}}=\frac{1}{a_{3}}-\frac{1}{a_{2}}=\cdots=\frac{1}{a_{11}}-\frac{1}{a_{10}}$ ), find the smallest possible value of $a_{1}$.
(2 marks)
設 $a_{1} 、 a_{2} 、 \cdots 、 a_{11}$ 為正整數, 其中 $a_{1}<a_{2}<\cdots<a_{11}$ 。若 $\frac{1}{a_{1}} 、 \frac{1}{a_{2}} 、 \cdots 、 \frac{1}{a_{11}}$ 成一等差數列（即 $\frac{1}{a_{2}}-\frac{1}{a_{1}}=\frac{1}{a_{3}}-\frac{1}{a_{2}}=\cdots=\frac{1}{a_{11}}-\frac{1}{a_{10}}$ )，求 $a_{1}$ 的最小可能值。
19. 2310
19. Let $M$ be the L.C.M. of $a_{1}, a_{2}, \ldots, a_{11}$, and set $m_{i}=\frac{M}{a_{i}}$. Then each $m_{i}$ is a positive integer and $m_{1}>m_{2}>\cdots>m_{11}$. Furthermore, $m_{1}, m_{2}, \ldots, m_{11}$ form an arithmetic sequence and $M$ is also the L.C.M. of $m_{1}, m_{2}, \ldots, m_{11}$. Write $b_{11}=b$ and $b_{i}-b_{i+1}=d$, where $b, d$ are positive integers. Note that $b$ and $d$ have to be relatively prime by the definition of $M$. We want to minimise
$$
a_{1}=\frac{\operatorname{lcm}(b, b+d, \ldots, b+10 d)}{b+10 d} .
$$

When $b=2$ and $d=1$, we find that $a_{1}=2310$. We prove below that $a_{1}$ cannot be any smaller, so that the answer would be 2310 . Indeed, if $d \geq 2$, then
$$
a_{1} \geq \frac{\operatorname{lcm}(b+5 d, b+6 d, \ldots, b+10 d)}{b+10 d}=\frac{(b+5 d)(b+6 d) \cdots(b+10 d)}{2 \cdot 2 \cdot 2 \cdot 3 \cdot(b+10 d)} \geq \frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3} d^{4}>2310 .
$$

Hence we may assume $d=1$. In this case
$$
a_{1} \geq \frac{\operatorname{lcm}(b+5, b+6, \ldots, b+10)}{b+10}=\frac{(b+5)(b+6) \cdots(b+10)}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot(b+10)} \geq \frac{(b+5)(b+6) \cdots(b+9)}{48}
$$

If $b \geq 4$, we see that $a_{1}>2310$. So it remains to check the cases $b=1,2,3$. Among these, we find that $b=2$ gives the minimum value $a_{1}=2310$.