## Task A-3.5.

For a natural number $n \geqslant 2$, let $D(n)$ be the greatest natural divisor of the number $n$ different from $n$. For example, $D(12)=6$ and $D(13)=1$.

Determine the greatest natural number $n$ such that $D(n)=35$.
## Solution.

Let $P(n)$ denote the smallest natural divisor of $n$ different from 1. Then,

$$
n = P(n) \cdot D(n) = 35 P(n)
$$

Also, $P(n)$ is a prime number.

Since 35 is a divisor of $n$, 5 is one of the prime factors of $n$, so $P(n)$ is less than or equal to 5, i.e., one of the numbers 2, 3, or 5.

The largest $n$ is obtained for $P(n) = 5$, i.e., for $n = 35 \cdot 5 = 175$.

Note: The answer $n = 175$ without additional explanation of why it is the largest such number earns 1 point.

Note: The first 2 points are not awarded for the claim that $n$ is a multiple of 35, i.e., $n = 35k$ for some $k$. It is necessary to highlight that $k$ is precisely the smallest natural divisor of $n$ different from 1, i.e., to use the fact that divisors come in pairs.