Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 12"
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==Solution== | ==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== | ==See also== |