2011 UNCO Math Contest II Problems/Problem 10

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.

2011 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2011 UNCO Math Contest II Problems/Problem 10

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