Eray wrote: A quadratic number is a real root of the equations $ax^2 + bx + c = 0$ where $|a|,|b|,|c|\in\{1,2,\ldots,10\}$. Find the smallest positive integer $n$ for which at least one of the intervals$$\left(n-\dfrac{1}{3}, n\right)\quad \text{and}\quad\left(n, n+\dfrac{1}{3}\right)$$does not contain any quadratic number. $x^2+x-3$ has a root $\frac{-1+\sqrt{13}}2\in\left(1,1+\frac 13\right)$ $x^2+3x-3$ has a root $\frac{-3+\sqrt{21}}2\in\left(1-\frac 13,1\right)$ So $1$ fits $x^2+x-7$ has a root $\frac{-1+\sqrt{29}}2\in\left(2,2+\frac 13\right)$ $x^2+x-5$ has a root $\frac{-1+\sqrt{21}}2\in\left(2-\frac 13,2\right)$ So $2$ fits $x^2-3x-1$ has a root $\frac {3+\sqrt{13}}2\in\left(3,3+\frac 13\right)$ $x^2-x-5$ has a root $\frac{1+\sqrt{21}}2\in\left(3-\frac 13,3\right)$ So $3$ fits For $n\in\{4,...,10\}$ : $x^2-nx-1$ has a root $\frac {n+\sqrt{n^2+4}}2\in\left(n,n+\frac 13\right)$ $x^2-nx+1$ has a root $\frac {n+\sqrt{n^2-4}}2\in\left(n-\frac 13,n\right)$ So $\{4,...,,10\}$ fit. The greatest possible quadratic number is $\frac{|b|+\sqrt{b^2+|4ac|}}{2|a|}$ whose maximum is obviously when $|b|=|c|=10$ and $|a|=1$ So this greatest is greatest root of $x^2-10x-10$, and so $\frac{10+\sqrt{140}}2<11$ So $11$ does not fit (no quadratic number in $\left(11,11+\frac 13\right)$) Hence the answer $\boxed{11}$