[ $\quad$ Completing the square. Sums of squares ]

For what values of $x$ and $y$ is the equality $x^{2}+(1-y)^{2}+(x-y)^{2}=1 / 3$ true?
The first method. After expanding the brackets and combining like terms, we get $0=2 x^{2}+2(1-y)^{2}+2(x-y)^{2}-2 / 3=4 x^{2}+4 y^{2}-4 x y-4 y+4 / 3=(2 x-y)^{2}+3(y-2 / 3)^{2}$. From here, $2 x=y=2 / 3$.

The second method. According to the Cauchy-Bunyakovsky inequality (see problem $\underline{61402}$ a) But the left side is also equal to 1, and the inequality turns into an equality when the vectors $(1,1,1)$ and $(x, 1-y, y-x)$ are proportional, that is, when $x=1-y=y-x \Leftrightarrow 2 x=y, 3 x=1$.

## Answer

At $x=1 / 3, y=2 / 3$.

Author: Emelyanova T.

Can the asterisks in the expression $\operatorname{HOK}(*, *, *)-\operatorname{HOK}(*, *, *)=2009$ be replaced with six consecutive natural numbers in some order so that the equality becomes true?

## Solution

Suppose such numbers were found. The least common multiple of several numbers is divisible by each of them and, consequently, by each of their divisors. Therefore, if there is an even number among the numbers from which the LCM is found, the LCM will also be even. The difference between two even numbers is an even number, but 2009 is odd. Therefore, all even numbers must be in one LCM, that is, one LCM will be found from three consecutive even numbers, and the other from three consecutive odd numbers. Among three consecutive even or odd numbers, there is a multiple of 3. Therefore, both LCMs are multiples of 3. But 2009 is not divisible by 3. Contradiction.

## Answer

No.