6. Find all integer values of $a$, not exceeding 15 in absolute value, for each of which the inequality

$$
\frac{4 x-a-4}{6 x+a-12} \leqslant 0
$$

is satisfied for all $x$ in the interval $[2 ; 3]$. In your answer, specify the sum of all such $a$.
Answer: -7.

Solution. Solving the inequality using the interval method, we find that its left side equals 0 at $x=1+\frac{a}{4}$ and is undefined at $x=2-\frac{a}{6}$. These two values coincide when $a=\frac{12}{5}$. In this case, the inequality has no solutions.

For $a>\frac{12}{5}$, the solution to the inequality is the interval $\left(2-\frac{a}{6} ; 1+\frac{a}{4}\right]$, and the inequality holds for all $x \in[2 ; 3]$ if $\left\{\begin{array}{c}2-\frac{a}{6}<2, \\ 3 \leqslant 1+\frac{a}{4}\end{array} \Leftrightarrow a \geqslant 8\right.$.

For $a<\frac{12}{5}$, the solution to the inequality is the interval $\left[1+\frac{a}{4} ; 2-\frac{a}{6}\right)$, and the inequality holds for all $x \in[2 ; 3]$ if $\left\{\begin{array}{l}1+\frac{a}{4} \leqslant 2, \\ 3<2-\frac{a}{6}\end{array} \Leftrightarrow a<-6\right.$.

Therefore, $a \in(-\infty ;-6) \bigcup[8 ;+\infty)$, and the required values are: $a \in[-15 ;-6) \bigcup[8 ; 15]$. The sum of the integer values:

$$
-15-14-13-\ldots-8-7+8+9+\ldots+14+15=-7
$$

## Set of Creative Problems.

I. Find all natural values of $n$ not exceeding 100 for which the sum $1^{2}+2^{2}+3^{2}+\ldots+n^{2}$ is divisible by 50. Arrange these values of $n$ in ascending order: $n_{1}<n_{2}<\ldots<n_{k}$. In the answer, specify $n_{4}$.

Solution. Since $1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$, then $n(n+1)(2 n+1)$ must be divisible by $300=2^{2} \cdot 3 \cdot 5^{2}$. Since the numbers $n, n+1$, and $2 n+1$ are pairwise coprime, one of them must be divisible by 25.

- If $n$ is divisible by 25, then $n=25,50,75$ or 100. Substituting these values into the expression, we find that divisibility holds for $n=75$ and $n=100$.
- If $n+1$ is divisible by 25, then $n=24,49,74$ or 99. Substituting these values into the expression, we find that divisibility holds for $n=24$ and $n=99$.
- If $2 n+1$ is divisible by 25 and is odd, then $n=12,37,62$ or 87. Substituting these values into the expression, we find that divisibility holds for $n=12$ and $n=87$.

Thus, the values that work are $12,24,75,87,99,100$. The fourth of these is 87.

Answer: 87.