Difference between revisions of "Chicken McNugget Theorem"
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==Proof 2== | ==Proof 2== | ||
− | |||
We start with this statement taken from [[Fermat%27s_Little_Theorem#Proof_2_.28Inverses.29|Proof 2 of Fermat's Little Theorem]]: | We start with this statement taken from [[Fermat%27s_Little_Theorem#Proof_2_.28Inverses.29|Proof 2 of Fermat's Little Theorem]]: | ||
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Clearly none of the <math>ia</math> for <math>1 \le i \le p-1</math> are divisible by <math>p</math>, so it suffices to show that all of the elements in <math>a \cdot S</math> are distinct. Suppose that <math>ai \equiv aj \pmod{p}</math> for <math>i \neq j</math>. Since <math>\text{gcd}\, (a,p) = 1</math>, by the cancellation rule, that reduces to <math>i \equiv j \pmod{p}</math>, which is a contradiction." | Clearly none of the <math>ia</math> for <math>1 \le i \le p-1</math> are divisible by <math>p</math>, so it suffices to show that all of the elements in <math>a \cdot S</math> are distinct. Suppose that <math>ai \equiv aj \pmod{p}</math> for <math>i \neq j</math>. Since <math>\text{gcd}\, (a,p) = 1</math>, by the cancellation rule, that reduces to <math>i \equiv j \pmod{p}</math>, which is a contradiction." | ||
− | Because <math>m</math> and <math>n</math> are coprime, we know that multiplying the residues of <math>m</math> by <math>n</math> simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form <math>am+bn</math>, <math>a</math> is <math>0</math> and <math>b</math> is the original residue. We now prove the following | + | Because <math>m</math> and <math>n</math> are coprime, we know that multiplying the residues of <math>m</math> by <math>n</math> simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form <math>am+bn</math>, <math>a</math> is <math>0</math> and <math>b</math> is the original residue. We now prove the following lemma. |
<b>Lemma</b>: For any nonnegative integer <math>c < m</math>, <math>cn</math> is the least purchasable number <math>\equiv cn \bmod m</math>. | <b>Lemma</b>: For any nonnegative integer <math>c < m</math>, <math>cn</math> is the least purchasable number <math>\equiv cn \bmod m</math>. | ||
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Putting it all together, we can say that for any coprime <math>m</math> and <math>n</math>, <math>mn-m-n</math> is the greatest number not representable in the form <math>am + bn</math> for nonnegative integers <math>a, b</math>. <math>\square</math> | Putting it all together, we can say that for any coprime <math>m</math> and <math>n</math>, <math>mn-m-n</math> is the greatest number not representable in the form <math>am + bn</math> for nonnegative integers <math>a, b</math>. <math>\square</math> | ||
+ | |||
+ | ==Corollary== | ||
+ | This corollary is based off of Proof 2, so it is necessary to read that Proof before this corollary. From the Lemma in Proof 2, every number <math>k \geq cn</math> that is congruent to <math>cn \bmod m</math> is purchasable. We prove the following lemma. | ||
+ | |||
+ | <b>Lemma</b> For any integer <math>k</math>, exactly one of the integers <math>k</math>, <math>mn-m-n-k</math> is not purchasable. | ||
+ | |||
+ | <i>Proof</i>: Because every number is congruent to some residue of <math>m</math> permuted by <math>n</math>, we can set <math>k \equiv cn \bmod m</math> for some <math>c</math>. We can break this into two cases. | ||
+ | |||
+ | <i>Case 1</i>: <math>k \leq cn-m</math>. This implies that <math>k</math> is not purchasable, and that <math>mn-m-n-k \geq mn-m-n-cn = n(m-1-c)</math>. <math>n(m-1-c)</math> is a permuted residue, and a result of the Lemma in Proof 2 showed that a permuted residue is the least number congruent to it <math>\bmod m</math> that is purchasable. Therefore, <math>mn-m-n-k \equiv n(m-1-c) \bmod m</math> and <math>mn-m-n-k \geq n(m-1-c)</math>, so <math>mn-m-n-k</math> is purchasable. | ||
+ | |||
+ | <i>Case 2</i>: <math>k >cn-m</math>. This implies that <math>k</math> is purchasable, and that <math>mn-m-n-k < mn-m-n-cn = n(m-1-c)</math>. Again, because <math>n(m-1-c)</math> is the least number congruent to it <math>\bmod m</math> that is purchasable, and because <math>mn-m-n-k \equiv n(m-1-c) \bmod m</math> and <math>mn-m-n-k < n(m-1-c)</math>, <math>mn-m-n-k</math> is not purchasable. | ||
+ | |||
+ | We now limit the values of <math>k</math> to all integers <math>0 \leq k \leq \frac{mn-m-n}{2}</math>, which limits the values of <math>mn-m-n-k</math> to <math>mn-m-n \geq mn-m-n-k \geq \frac{mn-m-n}{2}</math>. Because <math>m</math> and <math>n</math> are coprime, only one of them can be a multiple of <math>2</math>. Therefore, <math>mn-m-n \equiv (0)(1)-0-1 \equiv -1 \equiv 1 \bmod 2</math>, showing that <math>\frac{mn-m-n}{2}</math> is not an integer and that <math>\frac{mn-m-n-1}{2}</math> and <math>\frac{mn-m-n+1}{2}</math> are integers. We can now set equivalent limits on the values of <math>k</math> and <math>mn-m-n-k</math> so that they cover all integers form <math>0</math> to <math>mn-m-n</math> without overlap: <math>0 \leq k \leq \frac{mn-m-n-1}{2}</math> and <math>\frac{mn-m-n+1}{2} \leq mn-m-n-k \leq mn-m-n</math>. There are <math>\frac{mn-m-n-1}{2}+1=\frac{(m-1)(n-1)}{2}</math> values of <math>k</math>, and each is paired with a value of <math>mn-m-n-k</math>, so we make <math>\frac{(m-1)(n-1)}{2}</math> different ordered pairs of the form <math>(k, mn-m-n-k)</math>. The coordinates of these ordered pairs cover all integers from <math>0</math> to <math>mn-m-n</math> inclusive, and each contains exactly one not-purchasable integer, so that means that there are <math>\frac{(m-1)(n-1)}{2}</math> different not-purchasable integers from <math>0</math> to <math>mn-m-n</math>. All integers greater than <math>mn-m-n</math> are purchasable, so that means there are a total of <math>\frac{(m-1)(n-1)}{2}</math> nonnegative integers that are not purchasable. | ||
+ | |||
+ | In other words, for every pair of coprime integers <math>m, n</math>, there are exactly <math>\frac{(m-1)(n-1)}{2}</math> nonnegative integers that cannot be represented in the form <math>am + bn</math> for nonnegative integers <math>a, b</math>. | ||
==Generalization== | ==Generalization== | ||
− | |||
If <math>m</math> and <math>n</math> are not coprime, then we can simply rearrange <math>am+bn</math> into the form | If <math>m</math> and <math>n</math> are not coprime, then we can simply rearrange <math>am+bn</math> into the form | ||
<cmath>\gcd(m,n) \left( a\frac{m}{\gcd(m,n)}+b\frac{n}{\gcd(m,n)} \right)</cmath> | <cmath>\gcd(m,n) \left( a\frac{m}{\gcd(m,n)}+b\frac{n}{\gcd(m,n)} \right)</cmath> |
Revision as of 18:53, 18 February 2016
The Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) states that for any two relatively prime positive integers , the greatest integer that cannot be written in the form
for nonnegative integers
is
.
A consequence of the theorem is that there are exactly positive integers which cannot be expressed in the form
. The proof is based on the fact that in each pair of the form
, exactly one element is expressible.
Contents
Origins
The story goes that the Chicken McNugget Theorem got its name because in McDonalds, people bought Chicken McNuggets in 9 and 20 piece packages. Somebody wondered what the largest amount you could never buy was, assuming that you did not eat or take away any McNuggets. They found the answer to be 151 McNuggets, thus creating the Chicken McNugget Theorem.
Proof 1
Definition. An integer will be called purchasable if there exist nonnegative integers
such that
.
We would like to prove that is the largest non-purchasable integer. We are required to show that (1)
is non-purchasable, and (2) every
is purchasable.
Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.
Lemma. Let be the set of solutions
to
. Then
for any
.
Proof: By Bezout's Lemma, there exist integers such that
. Then
. Hence
is nonempty. It is easy to check that
for all
. We now prove that there are no others. Suppose
and
are solutions to
. Then
implies
. Since
and
are coprime and
divides
,
divides
and
. Similarly
. Let
be integers such that
and
. Then
implies
We have the desired result.
Lemma. For any integer , there exists unique
such that
.
Proof: By the division algorithm, there exists such that
.
Lemma. is purchasable if and only if
.
Proof: If , then we may simply pick
so
is purchasable. If
, then
if
and
if
, hence at least one coordinate of
is negative for all
. Thus
is not purchasable.
Thus the set of non-purchasable integers is . We would like to find the maximum of this set.
Since both
are positive, the maximum is achieved when
and
so that
.
Proof 2
We start with this statement taken from Proof 2 of Fermat's Little Theorem:
"Let . Then, we claim that the set
, consisting of the product of the elements of
with
, taken modulo
, is simply a permutation of
. In other words,
![\[S \equiv \{1a, 2a, \cdots, (p-1)a\} \pmod{p}.\]](http://latex.artofproblemsolving.com/9/e/3/9e3fcc9c2595bb1f400d5eeda2f8b487a50c4d5c.png)
Clearly none of the for
are divisible by
, so it suffices to show that all of the elements in
are distinct. Suppose that
for
. Since
, by the cancellation rule, that reduces to
, which is a contradiction."
Because and
are coprime, we know that multiplying the residues of
by
simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form
,
is
and
is the original residue. We now prove the following lemma.
Lemma: For any nonnegative integer ,
is the least purchasable number
.
Proof: Any number that is less than and congruent to it
can be represented in the form
, where
is a positive integer. If this is purchasable, we can say
for some nonnegative integers
. This can be rearranged into
, which implies that
is a multiple of
(since
). We can say that
for some positive integer
, and substitute to get
. Because
,
, and
. We divide by
to get
. However, we defined
to be a positive integer, and all positive integers are greater than or equal to
. Therefore, we have a contradiction, and
is the least purchasable number congruent to
.
This means that because is purchasable, every number that is greater than
and congruent to it
is also purchasable (because these numbers are in the form
where
). Another result of this Lemma is that
is the greatest number
that is not purchasable.
, so
, which shows that
is the greatest number in the form
. Any number greater than this and congruent to some
is purchasable, because that number is greater than
. All numbers are congruent to some
, and thus all numbers greater than
are purchasable.
Putting it all together, we can say that for any coprime and
,
is the greatest number not representable in the form
for nonnegative integers
.
Corollary
This corollary is based off of Proof 2, so it is necessary to read that Proof before this corollary. From the Lemma in Proof 2, every number that is congruent to
is purchasable. We prove the following lemma.
Lemma For any integer , exactly one of the integers
,
is not purchasable.
Proof: Because every number is congruent to some residue of permuted by
, we can set
for some
. We can break this into two cases.
Case 1: . This implies that
is not purchasable, and that
.
is a permuted residue, and a result of the Lemma in Proof 2 showed that a permuted residue is the least number congruent to it
that is purchasable. Therefore,
and
, so
is purchasable.
Case 2: . This implies that
is purchasable, and that
. Again, because
is the least number congruent to it
that is purchasable, and because
and
,
is not purchasable.
We now limit the values of to all integers
, which limits the values of
to
. Because
and
are coprime, only one of them can be a multiple of
. Therefore,
, showing that
is not an integer and that
and
are integers. We can now set equivalent limits on the values of
and
so that they cover all integers form
to
without overlap:
and
. There are
values of
, and each is paired with a value of
, so we make
different ordered pairs of the form
. The coordinates of these ordered pairs cover all integers from
to
inclusive, and each contains exactly one not-purchasable integer, so that means that there are
different not-purchasable integers from
to
. All integers greater than
are purchasable, so that means there are a total of
nonnegative integers that are not purchasable.
In other words, for every pair of coprime integers , there are exactly
nonnegative integers that cannot be represented in the form
for nonnegative integers
.
Generalization
If and
are not coprime, then we can simply rearrange
into the form
and
are coprime, so we apply Chicken McNugget to find a bound
We can simply multiply
back into the bound to get
Therefore, all multiples of
greater than
are representable in the form
for some positive integers
.
Problems
Introductory
- Marcy buys paint jars in containers of
and
. What's the largest number of paint jars that Marcy can't obtain?
- Bay Area Rapid food sells chicken nuggets. You can buy packages of
or
. What is the largest integer
such that there is no way to buy exactly
nuggets? Can you Generalize ?(ACOPS)
Intermediate
- Ninety-four bricks, each measuring
are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes
or
or
to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? Source
Olympiad
- On the real number line, paint red all points that correspond to integers of the form
, where
and
are positive integers. Paint the remaining integer point blue. Find a point
on the line such that, for every integer point
, the reflection of
with respect to
is an integer point of a different colour than
. (India TST)
- Let
be a set of integers (not necessarily positive) such that
(a) there exist with
;
(b) if and
are elements of
(possibly equal), then
also belongs to
.
Prove that is the set of all integers. (USAMO)