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

A. $8$

B. $0$

C. $11$

D. $5$

E. $1$

Solution

$f(28,17) = f(11,17) = f(6,11) = f(5,6) = f(1,5) = f(4,1) = f(3,1) = f(2,1) = f(1,1) = 1\ \mathrm{(E)}$

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