2006 SMT/General Problems/Problem 16

Solution

The condition we are looking for is $35(n-1)\equiv 0 \mod 360$ , because if any other $A_n = A_1$ than $35(n-1)$ must be a multiple of 360 as we have rotated $(n-1)$ by the time we place $A_n$ .

$35(n-1)\equiv 0\mod 360 \Rightarrow 5(n-1)\equiv 0\mod 360 \Rightarrow x\equiv 73,1\mod 360$

We can ignore the case of $n=1$ because be are looking for an $n\neq1$ . Therefore, our answer is the second lowest option, $n=73$ or $\boxed{A_{73}}$

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2006 SMT/General Problems/Problem 16

Solution