3. Some natural numbers form an increasing geometric progression with an integer common ratio. Find these numbers if their sum is 157.
Solution: If we consider that one number forms an increasing geometric progression, then the number 157 is a solution to the problem. According to the condition, for some natural number $n$, the equality $b_{1}+b_{1} q+b_{1} q^{2}+\cdots+b_{1} q^{n-1}=157$ holds, or $b_{1}\left(1+q+q^{2}+\cdots+q^{n-1}\right)=157$. Since 157 is a prime number (it is not divisible by $2,3,5,7,11$, 13), then $b_{1}=1,1+q+q^{2}+\cdots+q^{n-1}=157$, or $b_{1}=157,1+q+q^{2}+\cdots+q^{n-1}=1$. The second case does not fit. Therefore, $b_{1}=1, q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=156$, $q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=2^{2} \cdot 3^{1} \cdot 13$. Let's consider the following cases.

1) If $n=2$, then $q=156$. The required numbers: 1,156.
2) If $n=3$, then $q^{2}+q-156=0$. Since $q$ is a natural number greater than one, then $q=12$. The required numbers: $1,12,144$.
3) If $n \geq 4$, then $q\left(1+q+q^{2}+\cdots+q^{n-2}\right)=156$. Since $q$ is a natural number greater than one, then $q^{3} \leq q^{n-1} \leq 156, q \leq 5$, and $q$ is a divisor of the number $156=2^{2} \cdot 3^{1} \cdot 13$. Therefore, $q$ can take the following values: $2,3,4$.
4) If $q=2$, then $1+2+2^{2}+\cdots+2^{n-2}=78,2^{n-1}-1=156$, or $2^{n-1}=157$, the equation has no natural solutions.
5) If $q=3$, then $1+3+3^{2}+\cdots+3^{n-2}=52, \frac{3^{n-1}-1}{2}=52,3^{n-1}=105$, the equation has no natural solutions.
6) If $q=4$, then $1+4+4^{2}+\cdots+4^{n-2}=39, \frac{4^{n-1}-1}{3}=39,4^{n-1}=118$, the equation has no natural solutions.

Answer: $\{157\},\{1 ; 156\}$ or $\{1 ; 12 ; 144\}$.