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

$\begin{aligned} f (28, 17) & = f (11, 17) \\ = f (6, 11) \\ = f (5, 6) \\ = f (1, 5) \\ = f (4, 1) \\ = f (3, 1) \\ = f (2, 1) \\ = f (1, 1) \\ = 1 (E) \end{aligned}$

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

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2006 Cyprus MO/Lyceum/Problem 12

Problem

Solution

See also