3. Some natural numbers form an increasing geometric progression with an integer common ratio. Find these numbers if their sum is 211.
Solution: If we consider that one number forms an increasing geometric progression, then the number 211 is a solution to the problem. According to the condition, for some natural number $n$, the equality $b_{1}+b_{1} q+b_{1} q^{2}+\cdots+b_{1} q^{n-1}=211$ holds, or $b_{1}\left(1+q+q^{2}+\cdots+q^{n-1}\right)=211$. Since 211 is a prime number (it is not divisible by 2, 3, 5, 7, 11, 13, 17), then $b_{1}=1, 1+q+q^{2}+\cdots+q^{n-1}=211$ or $b_{1}=211, 1+q+q^{2}+\cdots+q^{n-1}=1$. The second case does not fit. Therefore, $b_{1}=1, q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=210$, $q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=2 \cdot 3 \cdot 5 \cdot 7$. Let's consider the following cases.

1) If $n=2$, then $q=210$. The required numbers: 1, 210.
2) If $n=3$, then $q^{2}+q-210=0$. Since $q$ is a natural number greater than one, then $q=14$. The required numbers: $1, 14, 196$.
3) If $n \geq 4$, then $q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=210$. Since $q$ is a natural number greater than one, then $q^{3} \leq q^{n-1} \leq 210, q \leq 5$, and $q$ is a divisor of the number $210=2 \cdot 3 \cdot 5 \cdot 7$. Therefore, $q$ can take the following values: $2, 3, 5$.
4) If $q=2$, then $1+2+2^{2}+\cdots+2^{n-2}=105, 2^{n-1}-1=105$, or $2^{n-1}=106$, the equation has no natural solutions.
5) If $q=3$, then $1+3+3^{2}+\cdots+3^{n-2}=70 \frac{3^{n-1}-1}{2}=70, 3^{n-1}=141$, the equation has no natural solutions.
6) If $q=5$, then $1+5+5^{2}+\cdots+5^{n-2}=42, \frac{5^{n-1}-1}{4}=42, 5^{n-1}=169$, the equation has no natural solutions.

Answer: $\{211\},\{1 ; 210\}$ or $\{1 ; 14 ; 196\}$.