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The number of hotdogs in a hotdog pack and the number of hotdog buns in a hotdog bun pack cannot be coprime
Steve Martin agrees!
(Although 8 and 12 aren’t coprime, and he tears open three bags of buns, meaning if he had just bought three packs of hot dogs and two bags of buns he’d be fine.)
Unless you get the (superior) Hebrew National dogs that come in packs of 7.
In which case, to get an even number of buns, he’d have to buy 12 packs of 7 hot dogs and 7 packs of 12 buns, for a total of 84 hot dogs and buns. (Or remove 5 buns from each 12 bun bag.)
Edit: Which means Joey Chestnut could eat his new record of 83 of those dogs in ten minutes and have another left over for Steve Martin!
“Coprime” is the operative qualifier of the original comment. You can’t do what Steve Martin did with coprime amounts of buns and dogs because they can never evenly go into one another. You’ll always have leftovers.
I did say that 8 and 12 weren’t coprime.
That isn’t true. You can do EXACTLY what he did. If he had packs of 8 hot dogs and 9 buns, removing one bun from each pack would have the same effect. And 8 and 9 are coprime.
And you can also do what I said he could’ve done, that is, get an even number of hot dogs and buns by purchasing different amounts of packages. If someone purchased 9 packs of 8 hot dogs and 8 packs of 9 buns, they would even out.
You can ensure any two coprime integers go into another number evenly by simply making them factors of the other number (in this case, 72).
Edit: fixed a typo
wish I could upvote you more than once just for the use of this word
We need more maths terminology in our regulations