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).