2006 Cyprus MO/Lyceum/Problem 12

Problem

If $f(\alpha,\beta)= \begin{cases}\alpha & \textrm {if} \qquad \alpha=\beta \\ f(\alpha-\beta,\beta) & \textrm {if} \qquad \alpha>\beta \\ f(\beta-\alpha,\alpha) & \textrm {if} \qquad \alpha<\beta \end{cases}$

then $f(28,17)$ equals

$\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1$

Solution

$\begin{align*} f(28,17)&=f(11,17)\\ &=f(6,11)\\ &=f(5,6)\\ &=f(1,5)\\ &=f(4,1)\\ &=f(3,1)\\ &=f(2,1)\\ &=f(1,1)\\ &=1 & \text{Thus the answer is}\mathrm{(E)} \end{align*}$

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 11	Followed by Problem 13
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2006 Cyprus MO/Lyceum/Problem 12

Problem

Solution

See also