Difference between revisions of "2011 UNCO Math Contest II Answer Key"

(Created page with "1) <math>41</math> 2) <math>\{22,28,30\}</math> 3) <math>65</math> 4) (a) <math>101</math> (b)<math> \{101, 325, 2501\}</math> 5) <math>160</math> 6) <math>9</math> 7)...")
 
 
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8) (a) <math>74</math>  (b) <math>45 \times 74</math>
 
8) (a) <math>74</math>  (b) <math>45 \times 74</math>
  
9) (a) <math>T(n+1)+T(n)=\binom{n}{3}</math>  (b) <math>T(N) = \binom{N 1}{3} − !\binom{N− 2}{3} + \binom{N 3}{3} \binom{N 4}{3} +\cdots</math>
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9) (a) <math>T(n+1)+T(n)=\binom{n}{3}</math>  (b) <math>T(N) = \binom{N-1}{3}-\binom{N-2}{3}+\binom{N-3}{3}-\binom{N-4}{3}+\cdots</math>
  
 
10) First try <math>\{1, 2, 3, \ldots , n\}</math> for <math>n= 2, 3, 4, 5</math>. The crossing off process yields <math>\{5,23,119,719\}</math> each one being one less
 
10) First try <math>\{1, 2, 3, \ldots , n\}</math> for <math>n= 2, 3, 4, 5</math>. The crossing off process yields <math>\{5,23,119,719\}</math> each one being one less
than a factorial. So for general <math>n</math> you should end up with<math>(n+ 1 )! 1</math>. Now look at <math>n=3</math> again and replace <math>1, 2, 3</math>
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than a factorial. So for general <math>n</math> you should end up with<math>(n+1)!-1</math>. Now look at <math>n=3</math> again and replace <math>1, 2, 3</math>
 
with <math>a,b,c</math> (order does not matter). Crossing off gives you
 
with <math>a,b,c</math> (order does not matter). Crossing off gives you
 
<cmath>(a+b+ab) + c + (a+b+ab)c  =a+b+c+ab+ac+bc+abc</cmath>
 
<cmath>(a+b+ab) + c + (a+b+ab)c  =a+b+c+ab+ac+bc+abc</cmath>
reminding one of the coefficients in
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reminding one of the coefficients in
 
<cmath>(x-a)(x-b)(x-c)= x^3-(a+b+c)x^2+(ab+ac+bc)x-abc</cmath>
 
<cmath>(x-a)(x-b)(x-c)= x^3-(a+b+c)x^2+(ab+ac+bc)x-abc</cmath>
Now let <math>x= −1</math>, and watch what happens remember that <math>\{a,b,c\} = \{1,2,3\} </math>.
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Now let <math>x=-1</math>, and watch what happens remember that <math>\{a,b,c\} = \{1,2,3\} </math>.
 
There are other approaches.
 
There are other approaches.
  
 
11) See solution to #2. Integers that are one less than a prime cannot be written in the form <math>m +n +m</math>.
 
11) See solution to #2. Integers that are one less than a prime cannot be written in the form <math>m +n +m</math>.

Latest revision as of 01:03, 6 November 2015

1) $41$

2) $\{22,28,30\}$

3) $65$

4) (a) $101$ (b)$\{101, 325, 2501\}$

5) $160$

6) $9$

7) $2^{502}-1504$

8) (a) $74$ (b) $45 \times 74$

9) (a) $T(n+1)+T(n)=\binom{n}{3}$ (b) $T(N) = \binom{N-1}{3}-\binom{N-2}{3}+\binom{N-3}{3}-\binom{N-4}{3}+\cdots$

10) First try $\{1, 2, 3, \ldots , n\}$ for $n= 2, 3, 4, 5$. The crossing off process yields $\{5,23,119,719\}$ each one being one less than a factorial. So for general $n$ you should end up with$(n+1)!-1$. Now look at $n=3$ again and replace $1, 2, 3$ with $a,b,c$ (order does not matter). Crossing off gives you \[(a+b+ab) + c + (a+b+ab)c  =a+b+c+ab+ac+bc+abc\] reminding one of the coefficients in \[(x-a)(x-b)(x-c)= x^3-(a+b+c)x^2+(ab+ac+bc)x-abc\] Now let $x=-1$, and watch what happens remember that $\{a,b,c\} = \{1,2,3\}$. There are other approaches.

11) See solution to #2. Integers that are one less than a prime cannot be written in the form $m +n +m$.

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