Nikolai with his son and Peter with his son went fishing. Nikolai caught as many fish as his son, and Peter caught three times as many as his son. In total, 25 fish were caught. What is the name of Peter's son?

#
Either three or four people went fishing. The question of the problem should clarify the situation.

## Solution

There are three possibilities: 1) four different people went fishing - Nikolai, Petr, and their sons; 2)

Petr is Nikolai's son; 3) Nikolai is Petr's son.

In the first case, each pair of father and son caught an even number of fish, which contradicts the condition.

In the second case, the total number of fish caught by Petr and Nikolai is six times greater than the number of fish caught by Petr's son, meaning the total number of fish must be divisible by 7. This also contradicts the condition.

Therefore, the only remaining possibility is that Nikolai is Petr's son. (Verify that this case does not contradict the condition.)

## Answer

Nikolai.

![](https://cdn.mathpix.com/cropped/2024_05_06_243bf411927c48073e50g-12.jpg?height=117&width=1175&top_left_y=0&top_left_x=11)
[ Algebraic inequalities (miscellaneous).]

Prove the inequality for natural numbers $n>1: \frac{(2 n)!}{(n!)^{2}}>\frac{4^{n}}{n+1}$.

## Solution

For $n=2,3$, the inequality can be verified directly, and for $n>3$,

$\frac{(2 n)!}{(n!)^{2}}=\frac{1}{1} \cdot \frac{3}{2} \cdot \frac{5}{3} \cdot \ldots \cdot \frac{2 n-1}{n} \cdot \frac{2}{1} \cdot \frac{4}{2} \cdot \frac{6}{3} \cdot \ldots \cdot \frac{2 n}{n} \cdot=$

$\frac{3}{1} \cdot \frac{5}{2} \cdot \frac{7}{3} \cdot \ldots \cdot \frac{2 n-1}{n-1} \cdot \frac{n+1}{n} \cdot 2^{n} \cdot \frac{1}{n+1}>\frac{4^{n}}{n+1}$, since $\frac{3}{1} \cdot \frac{5}{2} \cdot \frac{7}{3}=\frac{35}{2}>2^{4}$,

and each of the $n-4$ factors $\frac{9}{4}, \frac{11}{5}, \ldots, \frac{2 n-1}{n-1}$ is greater than 2.

Submit a comment