Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 10"

(Created page with "== Problem == The integers <math>1, 2, 3,\cdots , 50</math> are written on the blackboard. Select any two, call them <math>m</math> and <math>n</math> and replace these two with...")
 
(Solution)
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
 
+
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> 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> 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> Now let <math>x=-1</math>, and watch what happens remember that <math>\{a,b,c\} = \{1,2,3\}</math>. There are other approaches.
  
 
== See Also ==
 
== See Also ==

Revision as of 02:23, 13 January 2019

Problem

The integers $1, 2, 3,\cdots , 50$ are written on the blackboard. Select any two, call them $m$ and $n$ and replace these two with the one number $m+n+mn$. Continue doing this until only one number remains and explain, with proof, what happens. Also explain with proof what happens in general as you replace $50$ with $N$. As an example, if you select $3$ and $17$ you replace them with $3 + 17 + 51 = 71$. If you select $5$ and $7$, replace them with $47$. You now have two $47$’s in this case but that’s OK.


Solution

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.

See Also

2011 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions