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2006 SMT/Algebra Problems/Problem 9: Difference between revisions

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Problem

Principal Skinner is thinking of two integers $m$ and $n$ and bets Superintendent Chalmers that he will not be able to determine these integers with a single piece of information. Chalmers asks Skinner the numerical value of $mn+13m+13n-m^2-n^2$. From the value of this expression alone, he miraculously determines both $m$ and $n$. What is the value of the above expression?

Solution

Let $f(m,n)=mn+13m+13n-m^2-n^2$. Let $a=26-m$ and $b=26-n$. Then,


$f(a,b)=(26-m)(26-n)+13(26-m)+13(26-n)-(26-m)^2-(26-n)^2$ $=mn+13m+13n-m^2-n^2=f(m,n)$.


Thus, for any given value of $f(m,n)$, we have another pair of integers $(a,b)$ such that $f(m,n)=f(a,b)$. The only way in which this is not the case is if $a=m$ and $b=n$, or $26-m=m\implies m=13$ and $26-n=n\implies n=13$. Therefore, the value of the expression is $f(13,13)=\boxed{169}$.

See Also

2006 SMT/Algebra Problems

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