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

(sol)
(Standardized answer choices; minor edits)
Line 4: Line 4:
 
then <math>f(28,17)</math> equals
 
then <math>f(28,17)</math> equals
  
A. <math>8</math>
+
<math>\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1</math>
 
 
B. <math>0</math>
 
 
 
C. <math>11</math>
 
 
 
D. <math>5</math>
 
 
 
E. <math>1</math>
 
  
 
==Solution==
 
==Solution==
<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>
+
<math>\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*}</math>
  
 
==See also==
 
==See also==

Revision as of 09:36, 27 April 2008

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

$\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1$

Solution

$\begin{align*}f(28,17)&=f(11,17)\\ &=f(6,11)\\ &=f(5,6)\\ &=f(1,5)\\ &=f(4,1)\\ &=f(3,1)\\ &=f(2,1)\\ &=f(1,1)\\ &=1\ \mathrm{(E)}\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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