Review question

# Given this definition of $F(n)$, can we find the value of $F(6000)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8168

## Solution

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).