The Chicken McNugget Phrase, Explained (2024)

The past month hasBTS Chicken McNugget mealhas beeneverywhere.

The collaboration between McDonald's and K-Pop group BTS launched in the US on May 26, 2021, and for a brief moment, it felt like BTS and McNuggets were inseparable. The BTS meal was sold out at almost every McDonald's. Artist Josiah Chua transformed the packagingthese brand shoes. And of course there was that McNugget from the BTS mealsold for almost DKK 100,000on eBay because it looked like an Among Us character, which really makes me think McDonald's should have styled the BTS McNuggets to look like BTS. Can you imagine how much someone would pay for a McNugget that looks like Jungkook? Never underestimate how much people will pay for it 1)food that resembles pop culturein 2)something BTS related.

The promotion is now over (at least in the US), but there's still a lot to learn about Chicken McNuggets. For example, it used to be mathematically impossible to order 43 McNuggets at a McDonald's in the United Kingdom. Why? Because of a mathematical theorem aptly called the Chicken McNugget Theorem.

In honor of the BTS Chicken McNugget meal, I will explain the Chicken McNugget Theorem and its proof. We will review:

• Where the phrase Chicken McNugget came from (Hint: 19th century)
• How to prove the Chicken McNugget Theorem
• How to Apply the Chicken McNugget Phrase to the BTS Meal

In Britain, McDonald's only sold Chicken McNuggets in packs of 6, 9 and 20 (and unfortunately none of them looked like Jungkook). So if you want 12 McNuggets, you can of course buy two packs of 6. On the other hand, you can't buy 11 nuggets at once if your only options are 6, 9, and 20: that's impossible.

Mathematicians wanted to answer this question:

What is the largest number of McNuggets, youcan notbuy in packs of 6, 9 and 20?

After putting their blood, sweat and tears into it, the mathematicians discovered that the answer is 43. You can't buy 43 gold nuggets with packs of 6, 9 and 20, but you can buy any amount greater than 43.

The Chicken McNugget version of the problem probably originated in 1990 on the online newsgroup rec.puzzles, and was printed in Ilan Vardi's book in 1991Computational Recreations in Mathematics. But the original version of the problem goes all the way back to the late 19th century, long before McDonald's even existed.

The problem is often called the "Frobenius problem", after the German mathematician Ferdinand Frobenius. He sometimes mentioned the problem in his lectures. It is generally believed that the problem was first solved by the drug-addicted mathematician Joseph John Sylvester, who was at least aware of the problem.to in 1882.

People studied the Frobenius problem almost a hundred years before the McNugget version came about, and wrote theorems and formulas around the problem. But now the McNugget version is on its way, so let's get to McProving the Chicken McNugget Theorem.

Before we prove McNugget's theorem, we need to use a few concepts from a part of mathematics calledmodular arithmetic. This sounds big and complicated, but we only need to understand modular arithmetic notation and three concepts: greatest common factor, relative prime, and multiplicative inverse.

Notation:If you divide 20 by 3, you get 6 with a remainder of 2. Modular arithmetic is one way to express this remainder. When I write this:

20≡2 (mod. 3)

What it really means is that twenty have onerestof 2 if you divide it by 3 (we also say that 20 iscongruentto 2 versus 3).

Another way to say this is that if 20≡2 (mod 3), this means 20 = 3k+2 for an integer k. 20 is a multiple of 3 plus a remainder of 2. In this case, k=6, the equation is where . In general:

a≡r(mod b) → a=kb+r for an integer k.

Greatest common denominator:You probably remember itgreatest common denominator(sometimes calledgreatest common factor). The greatest common divisor of a and b is the largest number that divides both a and b. We write this as gcd(a,b). For example gcd(4,6)=2.

Relative prime:For many pairs of numbers the gcd is 1. For example, 3 and 7 have no common factors except 1, so gcd(3,7)=1. If two numbers have a gcd of 1, we call them numbersrelatively primaryofcoprime.

Multiplicative inverse:So let's say that ab≡1(mod m) for some numbers a, b and m. If the product ab is congruent with 1 mod m, we say that a and bmultiplicative inversefrom each other (against m). For example 10≡3(mod 7) and 5≡5(mod 7). 10*5 equals 50, and 50≡1(mod 7), so 10 and 5 are multiplicative inverses of 7.

This will be important because, without going into too much detail, we know that a multiplicative inverse of-inmust exist (against m)if a and m are relatively prime. For example, 9 and 5 are relatively prime numbers. That's why we know there is someBsuch that 9b=1 (mod 5). We know that 9 must have a multiplicative inverse of 5 (in this case the inverse of 9 is 4).

Let's say McDonald's sells two sizesgold nuggets. Name these measurements-inInB. Also assume that a and b are relatively prime, which means that gcd(a,b)=1. The Chicken McNugget sentence says the following:

Chicken McNugget-zin:For relatively prime numbers, positive integers-inInB, number oneM=ab-a-bcannot be expressed asax+woodfor all non-negative integersXInj. Also any integern>Mcan be represented asax+woodfor some non-negative integersXInj.

In other words, you cannot combine packagesin-size andB-size McNuggets to buy (ab-a-b)McNuggets, but you can buy any number of McNuggets greater than (ab-a-b). If there exist non-negative integers x,y such that ax+by=d, we will say that d isbe purchased. If they don't exist, we say they existunpurchasable.

Proving the Chicken McNugget theorem consists of two parts:

1. Prove that ab-a-b is thatunpurchasable.
2. Prove that every n>ab-a-b iscan be purchased,which means that there exists x,y≥0 such that xa+by=n.

Requirement 1: ab-a-b cannot be purchased.

We will use a proof of contradiction. We start by assuming the opposite of the statement: suppose ab-a-bbe purchased. If ab-a-b can be purchased, this means that there exist non-negative numbers x and y, such that ax+by=ab-a-b.

Als we (ax+with) delen door-in, we get a remainder ofdoor. So (axe+door)≡door(against a). In the same way (ab-a-b)≡ -B(mod-in). Since ax+by=ab-a-b this in turn implies thatdoor≡ -B(mod-in).

Since b and a are relatively prime, b must have a multiplicative inverse of a. Let us call this inverse b'. If we multiply both sidesdoor≡ -B(mod-in) at the inverse b' this happens:

b'door≡ b'-b(modin)

(b'b)y≡ -1(b'b)(mod -in)

j≡ -1(mod -in)

Similarly, if we divide both sides of the original equation (ax+by=ab-a-b) by b, we can reason that x≡-1 (mod b).

We know that x and y cannot be equal to -1, because we specified that they arenon-negative(You can't buy "-1 packs" of chicken nuggets). So the smallest possible value of x is b-1, and the smallest possible value of y is a-1. In other words: x≥b-1 and y≥a-1.

But if x≥b-1 and y≥a-1 this leads us to the following:

ab-a-b=bijl+door ≥ a(b-1)+b(a-1)= ab-a+ab-b= 2ab-a-b > ab-a-b.

This would mean that ab-a-b > ab-a-b, which is clearly a contradiction. Since being ab-a-bbe purchasedleads to a contradiction, we know that ab-a-b must be like thisunpurchasable.

Claim 2: Any n > ab-a-b can be purchased.

We now know that ab-a-b is not purchasable, but we also want to show that ab-a-b is.largestnon-purchasable number.

This proof becomes more complicated than Theorem 1. To begin, we must invoke the name of a dead Frenchman. That's right, it's about timeBit isidentity of salt:

Bézout's identity:If-inInBis an integer, then integers existx,ylike thatax+door=ggd(a,b).

In this case gcd(a,b)=1, so there exist some integers (x,y) such that ax+by=1. By multiplying both sides of the equation by n, we get nax+nby=n.

Let's define x'=nx and y'=ny to make our formula a bit clearer: ax'+by'=n. Now I would like to make another claim against you:

Lemma:If (x',y') is a solution to ax'+by'=n, then (x'-kb, y'+ka) is also a solution to any integer k.

Bevis:a(x’-kb)+b(y’+ka) = ax’-akb+door’+akb = ax’+door’=n.

Looking at the number x'-kb, k refers to the number of multiples of b that we add or subtract from x'. Note that we can choose a specific value of k such that x'-kb is between 0 and b-1 (0≤ x'-kb ≤b-1). Let's call this specific value k' and we define x"=x'-k'b and y"=y'+k'a. From the lemma we know that (x”,y”) is a solution for ax+by=n. But we're not quite done yet.We still need to prove that x" and y" are non-negative.

x" is by definition non-negative: we defined it as 0≤ x" ≤b-1. Now we need to show that y" is non-negative. We know that n=ax"+by", and n>ab-a-b. If you put them together, this is what happens:

ax”+by” > ab-a-b →

af”+b > ab-a-ax” →

b(y"+1) > a(b-1-x")

We know that x"≤b-1, which means that b-1-x"≥0. Number one-inis positive, so this also means a(b-1-x”)≥0, which means b(y”+1)>0. SinceBis positive, (y”+1)>0 and therefore y”≥0. We have now shown that x" and y" are non-negative, meaning that the numberNis available for purchase. QED.

McDonald's only sold the BTS Chicken McNugget meal in packs of 10. But let's say McDonald's decides to run the promotion again and also sells a 7-piece meal, because BTS has 7 members.

If the BTS Chicken McNugget meal comes in packs of 7 and 10, you cannot purchase 12 nuggets at a time or 15 nuggets. On the other hand, you can buy 17 or 21 nuggets in packs of 7 and 10. What is the largest number of BTS nuggets youcan notkopen?

To answer that we have the Chicken McNugget phrase. The largest number that cannot be purchased is:

ab-a-b = (7*10)–7–10 = 53.

In other words, you can't buy 53 nuggets at once, but you can buyeveryonenumber greater than 53.

Here are some other things we know about the Chicken McNugget phrase.

Why must a and b be relatively prime?Because otherwise there would be an infinite numberunpurchasablequantities of McNuggets. A simple example: in the US, Chicken McNuggets are sold in sizes 4 and 6 (and 10 and 20). The numbers 4 and 6 are not coprime because gcd(4,6)=2. It's easy to see that you can't buy an odd number of McNuggets with packs of 4 and 6. There is no maximum unpurchasable quantity.

In general, if gcd(a,b)=d, you can only buyNMcNuggets if n≡0(mod d). In other words, you can only buy n McNuggets if n is divisible by d. For example, if gcd(a,b)=5, then you cannot buy any amount that has a remainder of 1,2,3 or 4 when you divide it by 5 (so a number like 31 would be prohibitively expensive since 31≡1 (mod 5)).

In short, Chicken McNugget's theorem is only interesting if a and b are coprime.

How about 3 McNugget sizes instead of 2?We reviewed the proof for 2 different McNugget sizes, but what about 3 sizes? In principle, there is no explicit formula (and one spring day in 1990, Frank J. Curtis published onebevisthat there is no explicit formula).

But there are many algorithms to find the largest non-buyable amount for 3 McNugget sizes, the most notable of which isThe van Rødseth algorithm(1978),Davison's algorithm(1994), and the algorithm of Killingbergtrø (2000) (describedherwhen Killingbergtrø's original article was only in Norwegian).

Is the Chicken McNugget Phrase Useful?In short, yes! If you like my explanation ofsexy firsts, I mention that "sexy prime" is just a funny term without many actual applications or consequences in mathematics. However, the phrase Chicken McNugget is actually quite useful.

IHet Diophantine Frobenius-probelemby Jorge L. Ramírez Alfonsin, the author explains that the problem is related to algebraic geometry, error correction codes, vector spaces and monomial curves (in short, it is related to many topics important to mathematicians).

The problem can also help us solve the time complexity ofShell species. Basically, there is a sorting algorithm called Shellsort, but we don't know how long it takes for the algorithm to run, at least not today. The exact time is still an unresolved issue. Chicken McNugget's theorem helps us get a better idea of ​​how long the algorithm takes.

The BTS Chicken McNugget meal is no longer with us for a while. But even though we can now only buy regular packaged Chicken McNuggets, streaming”Butter' and longing for an earlier time, we know from the Chicken McNugget sentence that a certain amount of McNuggets was always not purchasable. There was always a certain amount of Chicken McNuggets from our McGrasp.

If there's a lesson to be learned from this, other than proving Chicken McNugget's theorem, it's that math can show how complicated something deep down really is. Even something as small and simple as a Chicken McNugget.