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If $f(0)=0, f(1)=1, f(2)=2$ and $f(x)=f(x-2)+f(x$ $-3)$ for $x=3,4,5 \ldots$, then $f(10)=$
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The correct answer is:
13
$\begin{aligned} \because & f(x)=f(x-2)+f(x-3) \\ & \therefore f(3)=f(1)+f(0)=1+0=1\end{aligned}$
$\begin{aligned} & f(4)=f(2)+f(1)=2+1=3 \\ & f(5)=f(3)+f(2)=1+2=3 \\ & f(6)=f(4)+f(3)=3+1=4 \\ & f(7)=f(5)+f(4)=3+3=6 \\ & f(8)=f(6)+f(5)=4+3=7 \\ & f(9)=f(7)+f(6)=6+4=10 \\ & f(10)=f(8)+f(7)=7+6=13\end{aligned}$
$\begin{aligned} & f(4)=f(2)+f(1)=2+1=3 \\ & f(5)=f(3)+f(2)=1+2=3 \\ & f(6)=f(4)+f(3)=3+1=4 \\ & f(7)=f(5)+f(4)=3+3=6 \\ & f(8)=f(6)+f(5)=4+3=7 \\ & f(9)=f(7)+f(6)=6+4=10 \\ & f(10)=f(8)+f(7)=7+6=13\end{aligned}$
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