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== | ||
equals
