Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 12"
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then <math>f(28,17)</math> equals | 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> | |
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− | B | ||
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− | C | ||
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− | D | ||
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− | E | ||
==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== |