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)...") |
|||
| Line 15: | Line 15: | ||
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 | + | 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 )! | + | 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 | |
<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= | + | 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 00:03, 6 November 2015
1) 
2) 
3) 
4) (a) 
(b)
5) 
6) 
7) 
8) (a) 
(b) 
9) (a) 
(b) 
10) First try 
for 
. The crossing off process yields 
each one being one less
than a factorial. So for general 
you should end up with
. Now look at 
again and replace 
with 
(order does not matter). Crossing off gives you
![\[(a+b+ab) + c + (a+b+ab)c =a+b+c+ab+ac+bc+abc\]](http://latex.artofproblemsolving.com/d/f/3/df32bf0089ce03bf07b1428033a04eff3c24eb46.png)
reminding one of the coefficients in
![\[(x-a)(x-b)(x-c)= x^3-(a+b+c)x^2+(ab+ac+bc)x-abc\]](http://latex.artofproblemsolving.com/9/7/c/97c99449bfa0c5bc6f98d38dc412d010d8d84de0.png)
Now let 
, and watch what happens remember that 
.
There are other approaches.
11) See solution to #2. Integers that are one less than a prime cannot be written in the form 
.
