The function \(F(n)\) is defined for all positive integers as follows; \(F(1)=0\), and for all \(n \geq 2\),
\[\begin{align*}
&F(n) = F(n-1) + 2 \quad\quad\quad\quad \text{if $2$ divides $n$ but $3$ does not divide $n$};\\
&F(n) = F(n-1) + 3 \quad\quad\quad\quad \text{if $3$ divides $n$ but $2$ does not divide $n$};\\
&F(n) = F(n-1) + 4 \quad\quad\quad\quad \text{if $2$ and $3$ both divide $n$};\\
&F(n) = F(n-1) \quad\quad\quad\quad\quad\quad \text{if neither $2$ nor $3$ divides $n$}.
\end{align*}\]
The value of \(F(6000)\) equals
\[(a)\quad 9827,\quad(b)\quad 10121,\quad(c)\quad 11000,\quad(d)\quad 12300,\quad(e)\quad 12352.\]
We know \(F(1) = 0\), and using the rules, \(F(2) = 2, F(3)=5, F(4) = 7, F(5) = 7, F(6) = 11\).
Thus as \(n\) travels from \(1\) to \(6\), we add on \(11\) to \(F(n)\).
But now the pattern repeats for \(7\) through to \(12\); we add on \(0\), then \(2\), then \(3\), then \(2\), then \(0\), then \(4\), which is \(11\).
This pattern will repeat for every \(6k+1, 6k+2, 6k+3, 6k+4, 6k+5, 6(k+1)\).
Thus we can split the numbers from \(1\) to \(6000\) into \(1000\) blocks of six, and the total added will be \(11000 = F(6000)\), and so the answer is (c).
