Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 12"

(Solution)
(Solution)
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==Solution==
 
==Solution==
\begin{align*}f(28,17)&=f(11,17)\\ &=f(6,11)\\ &=f(5,6)\\ &=f(1,5)\\
+
\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\ \mathrm{(E)}\end{align*}
+
&=f(4,1)\\ &=f(3,1)\\ &=f(2,1)\\ &=f(1,1)\\ &=1\ \mathrm{(E)}\end{align}
  
 
==See also==
 
==See also==

Revision as of 16:41, 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\ \mathrm{(E)}\end{align}

See also

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