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)\\ | |
| − | &=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 15:40, 23 November 2016
Problem
If 
then 
equals
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
