Difference between revisions of "Chicken McNugget Theorem"
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<math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math> [[2015 AMC 10B Problems/Problem 15|AMC 10B 2015 Problem 15]] | <math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math> [[2015 AMC 10B Problems/Problem 15|AMC 10B 2015 Problem 15]] | ||
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===Intermediate=== | ===Intermediate=== |
Revision as of 11:59, 13 July 2020
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
There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.
Proof Without Words
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 one and only one 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. 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 was that a permuted residue is the least number congruent to itself
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 itself
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 limits that are equivalent to the previous 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 can 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
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 relatively prime, then we can simply rearrange
into the form
and
are relatively prime, 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
Simple
- Marcy buys paint jars in containers of
and
. What's the largest number of paint jars that Marcy can't obtain?
Answer: containers
- 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)
Answer:
- If a game of American Football has only scores of field goals (
points) and touchdowns with the extra point (
points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?
Answer: points
- The town of Hamlet has
people for each horse,
sheep for each cow, and
ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
Answer:
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? AIME
- Find the sum of all positive integers
such that, given an unlimited supply of stamps of denominations
and
cents,
cents is the greatest postage that cannot be formed. AIME II 2019 Problem 14
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 points 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)