Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 12"
(New page: ==Problem== {{empty}} ==Solution== {{solution}} ==See also== {{CYMO box|year=2006|l=Lyceum|num-b=11|num-a=13}}) |
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| − | ==Problem== | + | == Problem == |
| − | {{ | + | If <math>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} </math> |
| + | |||
| + | then <math>f(28,17)</math> equals | ||
| + | |||
| + | <math>\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | <cmath>\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*}</cmath> | ||
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=11|num-a=13}} | {{CYMO box|year=2006|l=Lyceum|num-b=11|num-a=13}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
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