Zhendarov R.G.

In a box, there are 100 balls: white, blue, and red. It is known that if 26 balls are drawn from the box without looking, there will definitely be 10 balls of the same color among them. What is the smallest number of balls that need to be drawn from the box, without looking, to ensure that there are 30 balls of the same color among them?
Let's prove that 66 balls are sufficient. Suppose among them there are no more than 29 of each color. Then there are no less than $66-2 \cdot 29=$ 8 balls of each color and no less than $66-29=17$ of any two colors. But then there is a set of 8, 9, 9, which contradicts the condition.

65 balls are insufficient, for example, for the color distribution 7, 46, 47.

![](https://cdn.mathpix.com/cropped/2024_05_06_e123f5c8ab3546621c69g-16.jpg?height=108&width=1361&top_left_y=229&top_left_x=0)

Petya wrote down several algebraic expressions, squared each of them, and added the results.

Could he have ended up with the expression $x^{2}+y^{2}+z^{2}+3 y+4 x+x z+1$?

## Solution

If the expression $x^{2}+y^{2}+z^{2}+3 y+4 x+x z+1$ is equal to the sum of the squares of several algebraic expressions, then for any $x, y,$ and $z$, its value is non-negative. However, for example, when $x=z=0, y=-1, x^{2}+y^{2}+z^{2}+3 y+4 x+x z+1=1-3+1<0$.

## Answer

No, he could not.