Given this definition of $F(n)$, can we find the value of $F(6000)$?

Given this definition of $F(n)$, can we find the value of $F(6000)$?

Question

The function

$F(n)$

is defined for all positive integers as follows;

$F(1)=0$

, and for all

$n \geq 2$

,

$$\begin{align*} &F(n) = F(n-1) + 2 \quad\quad\quad\quad \text{if }2\text{ divides }n\text{ but }3\text{ does not divide }n\text{};\\ &F(n) = F(n-1) + 3 \quad\quad\quad\quad \text{if }3\text{ divides }n\text{ but }2\text{ does not divide }n\text{};\\ &F(n) = F(n-1) + 4 \quad\quad\quad\quad \text{if }2\text{ and }3\text{ both divide }n\text{};\\ &F(n) = F(n-1) \quad\quad\quad\quad\quad\quad \text{if neither }2\text{ nor }3\text{ divides }n\text{}. \end{align*}$$

The value of $F(6000)$ equals

$$(a)\quad 9827,\quad(b)\quad 10121,\quad(c)\quad 11000,\quad(d)\quad 12300,\quad(e)\quad 12352.$$

Suggestion

Can we use the rules to evaluate $F(n)$ for small $n$?

Can we find any pattern in the sequence given by the amounts we add on each time? Might this sequence be periodic?

Solution

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).