3. Find the smallest natural number $N$, which is divisible by $r$, ends in $r$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and $2p+1$ is the cube of a natural number.
Answer: 11713.

Solution. Let $2 p+1=n^{3}$. Then $(n-1)\left(n^{2}+n+1\right)=2 p$. The number $2 p$ can only have the following positive divisors: $1,2, p, 2 p$. The number $n$ is obviously odd, so $n-1$ is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=p$. From this, $n=3, p=13$.

The desired number $N$ can be represented as $N=X \cdot 100+13$, where the number $X$ is the number $N$ without the last two digits. Since $N$ is divisible by 13, then $X$ is also divisible by 13. The sum of the digits of the number $X$ is $13-(1+3)=$ 9, so $X$ is divisible by 9, and since 9 and 13 are coprime, $X$ is divisible by $9 \cdot 13=117$.

The smallest of such numbers is the number $X=117$, then $N=117 \cdot 100+13=11713$.

Comment. A correct and justified solution - 7 points. The value found by trial and not shown to be unique - 1 point; the value found and its uniqueness proven - 3 points. If the value of $N$ is found by trial but not shown to be the smallest, 1 point is added. If the solution is not completed, 1-2 points are awarded for proving useful auxiliary statements.