Difference between revisions of "2014 UNCO Math Contest II Answer Key"

m (Created page with "1) The youngest will be <math>16</math> and the perfect square is <math>100</math>. 2) <math>145</math> 3) <math>x =-2</math> and <math>y = \tfrac{-3}{4}</math> 4) (a) <mat...")
 
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4) (a) <math>210</math>    (b) <math>4010</math>    (c) <math>\tfrac{n(n^2+1)}{2}</math>
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4) (a) <math>210</math>    (b) <math>4010</math>    (c) <math>\frac{n(n^2+1)}{2}</math>
  
  
5) (a) <math>\tfrac{1}{5}</math>    (b) <math>\tfrac{(1-x)^2}{1+x^2}</math>
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5) (a) <math>\frac{1}{5}</math>    (b) <math>\frac{(1-x)^2}{1+x^2}</math>
  
  
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7) <math>\ftrac{112}{125}</math>
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7) <math>\frac{112}{125}</math>
  
  
8) (a) <math>\tfrac{2}{5}</math>    (b) <math>\tfrac{8}{31}</math>
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8) (a) <math>\frac{2}{5}</math>    (b) <math>\frac{8}{31}</math>
  
  

Latest revision as of 02:55, 22 October 2014

1) The youngest will be $16$ and the perfect square is $100$.


2) $145$


3) $x =-2$ and $y = \tfrac{-3}{4}$


4) (a) $210$ (b) $4010$ (c) $\frac{n(n^2+1)}{2}$


5) (a) $\frac{1}{5}$ (b) $\frac{(1-x)^2}{1+x^2}$


6) (a) $11$ (b) $76$


7) $\frac{112}{125}$


8) (a) $\frac{2}{5}$ (b) $\frac{8}{31}$


9) Zero probability for $n=1$ and $\frac{F_{n-1}}{2^{2n}}$ for $n\ge 2$, where $F_{n-1}$ is the $(n-1)^{st}$ Fibonacci number.


10)(a) $\tfrac{1}{11}$ (b) $\tfrac{2}{11}$ (c) $\tfrac{1023}{5120} = \left(\tfrac{1}{m+1}\right) \tfrac{2^{m+1}-1}{2^m}$,

where there are $m$ guessers, not including Tweedledee and Tweedledum. Here $m = 9$.