**The Chicken Nugget Puzzle**

McDonald’s sells chicken nuggets in packs of 4, 6, and 9 (I’m excluding 20 because mathematically, it’s irrelevant here).

If the employees can only sell chicken nuggets in these amounts, what is the LARGEST number of chicken nuggets that you can order that they WON’T be able to serve you (These numbers are called Frobenius Numbers).

Well… The only way I can think to solve this problem is by seeing if it’s possible or not

You **can’t** buy 1, 2, or 3 nuggets because that won’t fill a box and it says you have to sell the full box.

You can sell 4 (that’s one box of 4).

You **can’t** sell 5.

You can 6 (Thats one box of 6).

You **can’t** sell 7.

You can sell 8 (That’s two boxes of 4).

You can sell 9 (one box of 9).

You can sell 10 (one box of 6 and one box of 4)

Yan **can’t **sell 11.

You can sell 12 (three of 4 or two of six)

You can sell 13 (one of 9 and one of 4)

You can sell 14 (one of 6 and two of 4)

You can sell 15 (one of 9 and one of 6)

You can sell 16 (two of 6 and one of 4)

…

**How do you know when you’re done?!?!**

To be honest… I don’t know how to solve it. I just realized that this math is *way too complicated for me.*

By the way – for the record, I’m not *great* at math – I just like it.

“Will I give up?” you ask.

So now, if you’d like to stick around, I’m going to jot down my internal struggle to solving this problem 🙂

I found this handy dandy little formula that will give you the largest amount of chicken nuggets if you only have two box amounts (once you start talking about three or more it gets complicated. maybe someday I’ll understand it).

Largest Frobenius Number=(box size)(other box size)-box size-other box size

And I realized… I may not be able to solve the problem using the complicated method, but I can figure out how if a number of nuggets is possible to sell or not. Which brings me back to the question **How do I know when I’m done?**

If you can only only sell two of these sizes, you won’t be able to sell certain amounts of chicken nuggets. BUT if you put in a third size, you’re guaranteed to not have less options of amounts to sell. So… the smallest number of the three largest Frobenius Numbers should be the last number of chicken nugget requests that we’d need to test in order to conclude that we figured out Frobenius Number of Chicken Nuggets.

Here are the largest Frobenius Numbers:

(9)(6)-9-6=39

(9)(4)-9-4=23

(6)(4)-6-4=14

So theoretically, if I can find a way to combine 9 6 and 4 to be able to sell 14 chicken nuggets or less, I can figure out the largest number of chicken nuggets that you can order to piss off your Mc’Donalds cashier .

Okay… but wait. 14 is not a Frobenius number for 6 and 4 because 6+4+4=14…

And 39 is not a Frobenius number for 9 and 6 because 6+6+6+6+6+9=39.

So… **Apparently**, Frobenius Numbers have to be calculated using sets of numbers that have a greatest common factor (gcf) of 1.

The factors of 4 are 1 2 and 4. The factors of 6 are 1 2 3 and 6. Their gcf is 2. So the formula doesn’t work, which was troublesome to me, but then I realized why it doesn’t work… The answer is infinitely large. If your gcf is 2 and you can only use 6 and 4 nugget boxes, you can’t sell ANY odd number of chicken nuggets (an even number plus an even number is an even number).

The factors of 6 are 1 2 3 and 6. The factors of 9 are 1 3 and 9. Their gcf is 3. So the formula doesn’t work, because you can’t make any values that aren’t a multiple of 3.

BUT the factors of 9 are 1 3 and 9. The factors of 4 are 1 2 and 4. Their gcf is 1! So the formula works!

SO LONG STORY SHORT… If we can decide if it’s possible to sell the amount of chicken nuggets ranging from 1 to 23, we will know our largest Frobenius Number

You **can’t** buy 1, 2, or 3 nuggets because that won’t fill a box and it says you have to sell the full box.

You can sell 4 (that’s one box of 4).

You **can’t** sell 5.

You can 6 (Thats one box of 6).

You **can’t** sell 7.

You can sell 8 (That’s two boxes of 4).

You can sell 9 (one box of 9).

You can sell 10 (one box of 6 and one box of 4)

Yan **can’t **sell 11.

You can sell 12 (two of six)

You can sell 13 (one of 9 and one of 4)

You can sell 14 (one of 6 and two of 4)

You can sell 15 (one of 9 and one of 6)

You can sell 16 (two of 6 and one of 4)

You can sell 17 (two of 4 and one of 9)

You can sell 18 (two of 9)

You can sell 19 (one of 6, one of 4 and one of 9)

You can sell 20 (five of 4)

You can sell 21 (two of 6 and one of 9)

You can sell 22 (two of 9 and one of 4)

You can sell 23 (one of 6, two of 4 and one of 9)

FINAL ANSWER: 11 nuggets

So when you go to Mc’Donalds, ask for 11 nuggets and you’ll either confuse people or get a free nugget.

Why did I do that? Because I like a challenge and I don’t like taking answers if I don’t understand them. Math is complicated and there are so many things we don’t know and there are so many algorithms that we could simplify… but it feels so flibbing good to search for an answer and find one.

I like that there are answers that we can find and proofs that we can give in math. Because, well there aren’t many subjects that you can study where the answer is black and white. And, even in math, sometimes it’s not black and white. But it’s amazing when you can find the answer and understand all of the work that you did.

Thank you for sticking around for that entire post. It was kinda heavy.