Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 12"

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==Problem==
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== Problem ==
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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>
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then <math>f(28,17)</math> equals
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A. <math>8</math>
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B. <math>0</math>
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C. <math>11</math>
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D. <math>5</math>
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E. <math>1</math>
  
 
==Solution==
 
==Solution==
{{solution}}
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<math>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)}</math>
  
 
==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]]

Revision as of 19:30, 17 October 2007

Problem

If $f(\alpha,\beta)= \begin{cases}\alpha & \textrm {if} \qquad \alpha=\beta \\ f(\alpha-\beta,\beta) & \textrm {if} \qquad \alpha>\beta \\ f(\beta-\alpha,\alpha) & \textrm {if} \qquad \alpha<\beta \end{cases}$

then $f(28,17)$ equals

A. $8$

B. $0$

C. $11$

D. $5$

E. $1$

Solution

$f(28,17) = f(11,17) = f(6,11) = f(5,6) = f(1,5) = f(4,1) = f(3,1) = f(2,1) = f(1,1) = 1\ \mathrm{(E)}$

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 11
Followed by
Problem 13
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