5. Find the smallest natural number $A$, which is divisible by $p$, ends in $\boldsymbol{p}$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and is the cube of a natural number.
Solution. Let $2 \mathrm{p}+1=\mathrm{n}^{3}$. Then $(\mathrm{n}-1)\left(\mathrm{n}^{2}+n+1\right)=2 \mathrm{p}$. The number $2 \mathrm{p}$ can only have the following positive divisors: 1, 2, p, 2p. The number n is clearly 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 A can be represented as $\mathrm{A}=\mathrm{B} \times 100+13$, where the number $\mathrm{B}$ is the number A without the last two digits. Since A is divisible by 13, then B is also divisible by 13. The sum of the digits of the number B is $13-(1+3)=9$, so B is divisible by 9, and since 9 and 13 are coprime, B is divisible by $9 \times 13=117$. The smallest such number is $B=117$, then $A=117 \times 100+13=11713$.

Answer: 11713.