2011 UNCO Math Contest II Problems/Problem 10
Contents
Problem
The integers 
are written on the blackboard. Select any two, call them 
and 
and replace
these two with the one number 
. 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 
with 
. As an example, if you select 
and 
you replace them with 
. If you select 
and 
, replace them with 
. You now have two ’s in this case but that’s OK.
Solution 1
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.
Solution 2
By using Simon's Favorite Factoring Trick, we can see that 
Something interesting happens when we set 
and 
\begin{align*}
(m+1)(n+1)-1&=\\
(a-1+1)(b-1+1)-1&=\\
ab-1
\end{align*}
If we represent each element of the set as 
the action repeatedly performed becomes much simpler.
For a set 
performing the action stated in the problem until one value is left yields the value 
Setting 
to the first 50 integers and perform the action until one value is left, we get a value equal to 
Setting to the first integers yeilds a value equal to 
See Also
| 2011 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
