Difference between revisions of "Chicken McNugget Theorem"

(Added proof. Needs some formatting.)
m
Line 9: Line 9:
 
Proof:
 
Proof:
 
Each member of the residue class can be written as  
 
Each member of the residue class can be written as  
<math>am + r</math> for an integer <math>a</math>. Since <math>r</math> is in the form <math>bn</math>, this can be rewritten as <math>am + br</math>.
+
<math>am + r</math> for an integer <math>a</math>. Since <math>r</math> is in the form <math>bn</math>, this can be rewritten as <math>am + bn</math>.
 
Nonnegative values of <math>a</math> correspond to members greater than or equal to <math>r</math>. Negative values of <math>a</math> correspond to members less than <math>r</math>. Thus the lemma is proven.
 
Nonnegative values of <math>a</math> correspond to members greater than or equal to <math>r</math>. Negative values of <math>a</math> correspond to members less than <math>r</math>. Thus the lemma is proven.
  

Revision as of 20:11, 22 July 2008

The Chicken McNugget Theorem states that for any two relatively prime positive integers $m,n$, the greatest integer that cannot be written in the form $am + bn$ for nonnegative integers $a, b$ is $mn-m-n$.

Proof

Consider the integers $\pmod{m}$. Let $R = \{0, n, 2n, 3n, 4n ... (m-1)n\}$. Note that since $m$ and $n$ are relatively prime, $R$ is a Complete residue system in modulo $m$.

Lemma: For any given residue class $S \pmod{n}$, call $r$ the member of $R$ in this class. All members greater than or equal to $r$ can be written in the form $am+bn$ while all members less than $r$ cannot for nonnegative $a,b$.

Proof: Each member of the residue class can be written as $am + r$ for an integer $a$. Since $r$ is in the form $bn$, this can be rewritten as $am + bn$. Nonnegative values of $a$ correspond to members greater than or equal to $r$. Negative values of $a$ correspond to members less than $r$. Thus the lemma is proven.

The largest member of $R$ is $(m-1)n$, so the largest unattainable score $p$ is in the same residue class as $(m-1)n$.

The largest member of this residue class less than $(m-1)n$ is $(m-1)n - m = mn - m - n$ and the proof is complete.

See Also

This article is a stub. Help us out by expanding it.