A sequence of positive integers $a_{1}, a_{2}, \ldots$ is such that for each $m$ and $n$ the following holds: if $m$ is a divisor of $n$ and $m<n$, then $a_{m}$ is a divisor of $a_{n}$ and $a_{m}<a_{n}$. Find the least possible value of $a_{2000}$.
Answer: 128.

Let $d$ denote the least possible value of $a_{2000}$. We shall prove that $d=128$.

Clearly the sequence defined by $a_{1}=1$ and $a_{n}=2^{\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k}}$ for $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots \cdots p_{k}^{\alpha_{k}}$ has the required property. Since $2000=2^{4} \cdot 5^{3}$, we have $a_{2000}=2^{4+3}=128$ and therefore $d \leqslant 128$.

Note that in the sequence $1<5<25<125<250<500<1000<2000$ each term is divisible by the previous one. As $a_{1} \geqslant 1$ it follows that

$$
a_{2000} \geqslant 2 \cdot a_{1000} \geqslant 2^{2} \cdot a_{500} \geqslant 2^{3} \cdot a_{250} \geqslant \ldots \geqslant 2^{7} \cdot a_{1} \geqslant 2^{7}=128 .
$$