Problem 11.7. Let $N$ be the least common multiple of ten different natural numbers $a_{1}<a_{2}<a_{3}<\ldots<a_{10}$.

(a) (2 points) What is the smallest value that $N / a_{3}$ can take?

(b) (2 points) Identify all possible values of $a_{3}$ in the interval $[1 ; 1000]$, for which the value of $N / a_{1}$ can take its smallest value.
Answer: (a) 8. (b) $315,630,945$.

Solution. (a) Note that $N$ is divisible by any of the numbers $a_{1}, a_{2}, \ldots, a_{10}$. This means that the numbers $a_{1}, a_{2}, \ldots, a_{10}$ can be represented as $a_{1}=N / b_{1}, a_{2}=N / b_{2}, \ldots, a_{10}=N / b_{10}$, where $b_{1}, b_{2}, \ldots, b_{10}$ are different natural numbers. Since $a_{1}b_{2}>\ldots>b_{10}$. It is clear from this that $N / a_{3}=b_{3} \geqslant 8$, since it is greater than some 7 different natural numbers.

At the same time, the value $b_{3}=8$ is achieved, for example, in the case of

$$
N=10!, a_{1}=\frac{10!}{10}, a_{2}=\frac{10!}{9}, a_{3}=\frac{10!}{8}, \ldots, a_{10}=\frac{10!}{1}
$$

(b) Similarly to part (a), we can understand that the smallest value of $b_{1}=N / a_{1}$ is 10, and it can only be achieved if $b_{10}=1, b_{9}=2, \ldots, b_{2}=9$. Note that the numbers $b_{1}, b_{2}, \ldots, b_{10}$ are divisors of the number $N$, i.e., $N$ must be divisible by all natural numbers from 1 to 10. Thus, $N=k \cdot$ LCM $(1,2, \ldots, 10)$.

To find the LCM of the numbers from 1 to 10, we need to determine for each prime number the highest power by which these numbers are divisible. For 2, this power is $3\left(8=2^{3}\right)$; for 3, it is $2\left(9=3^{2}\right)$; for 5 and 7, it is 1. Thus,

$$
\text { LCM }(1,2, \ldots, 10)=2^{3} \cdot 3^{2} \cdot 5 \cdot 7=2520
$$

From this, we get that

$$
a_{3}=\frac{N}{b_{3}}=\frac{N}{8}=\frac{k \cdot 2520}{8}=315 \cdot k
$$

Among the numbers from 1 to 1000, exactly 3 numbers have the required form: $315,630,945$. All of them satisfy the condition of the problem.