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(D) Given the equation $[x]\{x\} = 5$ where $x \in [0, 2015]$.Let $n = [x]$ and $f = \{x\}$,where $n$ is an integer and $0 \leq f < 1$.The equation be

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$2009$

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(D) Given the equation $[x]\{x\} = 5$ where $x \in [0, 2015]$.Let $n = [x]$ and $f = \{x\}$,where $n$ is an integer and $0 \leq f The equation becomes $n \cdot f = 5$,which implies $f = \frac{5}{n}$.Since $0 \leq f This implies $n > 5$ (since $n$ must be positive for $f$ to be positive).Also,$x = n + f = n + \frac{5}{n}$.Given $x \leq 2015$,we have $n + \frac{5}{n} \leq 2015$.Since $n$ is an integer and $n > 5$,the possible values for $n$ are $6, 7, \dots, 2014$.For each such $n$,$x = n + \frac{5}{n}$ is a valid solution because $0 \leq \frac{5}{n}  5$.If $n = 2015$,$x = 2015 + \frac{5}{2015} > 2015$,which is outside the range.Thus,$n$ can take values from $6$ to $2014$.The number of such values is $2014 - 6 + 1 = 2009$.

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$,and let $\{x\} = x - [x]$. The number of solutions $x$ to the equation $[x]\{x\} = 5$ with $0 \leq x \leq 2015$ is

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