If $\left|-1007x+2013\right|<2015$, how many integers could $x$ be?

เฉลยละเอียด

[STEP]Simplify the inequality[/STEP]

\begin{eqnarray*}
\left|-1007x+2013\right| & < & 2015\\
\left|1007x-2013\right| & < & 2015\\
\left|x-\frac{2013}{1007}\right| & < & \frac{2015}{1007}
\end{eqnarray*}

[STEP]Using the definition of absolute values to separate it into two inequalities[/STEP]

\begin{eqnarray*}
\frac{2013}{1007}-\frac{2015}{1007} & <x< & \frac{2013}{1007}+\frac{2015}{1007}\\
-\frac{2}{1007} & <x< & \frac{4028}{1007}
\end{eqnarray*}

The integers satifying the inequality are $0$, $1$, $2$ and $3$. [ANS]So there are $4$ of them[/ANS]

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