3. Some natural numbers form an increasing geometric progression with an integer common ratio. Find these numbers if their sum is 463.
Solution: If we consider that a single number forms an increasing geometric progression, then the single number 463 is a solution to the problem. Let the progression have no fewer than two terms. According to the condition, for some natural number $n \geq 2$, the equality $b_{1}+b_{1} q+b_{1} q^{2}+\cdots+b_{1} q^{n-1}=463$ holds, or $b_{1}\left(1+q+q^{2}+\cdots+q^{n-1}\right)=463$. Since 463 is a prime number (it is not divisible by $2,3,5,7,11$, $13,17,19,23$), then $b_{1}=1,1+q+q^{2}+\cdots+q^{n-1}=463$, or $b_{1}=463,1+q+q^{2}+\cdots+q^{n-1}=1$. The second case is not suitable. Therefore, $b_{1}=1, q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=462$, $q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=2 \cdot 3 \cdot 7 \cdot 11$. Consider the following cases.

1) If $n=2$, then $q=462$. The required numbers are: 1,462.
2) If $n=3$, then $q^{2}+q-462=0$. Since $q$ is a natural number greater than one, then $q=21$. The required numbers are: $1,21,441$.
3) If $n \geq 4$, then $q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=462$. Since $q$ is a natural number greater than one, then $q^{3} \leq q^{n-1} \leq 462, q \leq 7$, and $q$ is a divisor of the number $462=2 \cdot 3 \cdot 7 \cdot 11$. Therefore, $q$ can take the following values: $2,3,6,7$.
4) If $q=2$, then $1+2+2^{2}+\cdots+2^{n-2}=231,2^{n-1}-1=231$, or $2^{n-1}=232$, the equation has no natural solutions.
5) If $q=3$, then $1+3+3^{2}+\cdots+3^{n-2}=154, \frac{3^{n-1}-1}{2}=154,3^{n-1}=309$, the equation has no natural solutions.
6) If $q=6$, then $1+6+6^{2}+\cdots+6^{n-2}=77, \frac{6^{n-1}-1}{5}=77,6^{n-1}=386$, the equation has no natural solutions.
7) If $q=7$, then $1+7+7^{2}+\cdots+7^{n-2}=66, \frac{7^{n-1}-1}{6}=66,7^{n-1}=397$, the equation has no natural solutions.

Answer: $\{463\},\{1 ; 462\}$ or $\{1 ; 21 ; 441\}$.