2006 Cyprus MO/Lyceum/Problem 12: Difference between revisions

Latest revision as of 15:51, 23 November 2016

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
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: Difference between revisions

Latest revision as of 15:51, 23 November 2016

Problem

Solution

See also

@@ Line 7: / Line 7: @@
 ==Solution==
-\begin{align}f(28,17)&=f(11,17)\\ &=f(6,11)\\ &=f(5,6)\\ &=f(1,5)\\
+<cmath>\begin{align*}
-&=f(4,1)\\ &=f(3,1)\\ &=f(2,1)\\ &=f(1,1)\\ &=1\ \mathrm{(E)}\end{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*}</cmath>
 ==See also==