43 McNuggets Made Difficult


I mentioned in the last comments thread the McNuggets Problem, and realized belatedly that maybe not everybody knew just what that was. It’s a cute little one, which Wolfram’s Mathworld is able to date to 1991, or maybe 1990. There’s a reference to a March 1990 puzzle on Usenet newsgroup rec.puzzles, but to find it would require some Google-like search engine capable of finding postings on Usenet, and that technology is sadly beyond us.

Whether 1990 or 1991 seems late, since I’m certain the puzzle first appeared about the same time people first saw the original Chicken McNuggets menu options on sale, sometime in the mid-80s. In the original offerings, one could buy a pack of six, nine, or if Mom was feeling particularly flush with cash or you gave a credible impersonation of being willing to share with your siblings, twenty. The obvious question, then, is what’s the largest number of McNuggets which can’t be bought by some combination of these?

This can be studied rigorously, although I don’t know anyone who actually would. It’s more fun to play and see what can be constructed: 12, obviously; 15, as surely; 18 as well (and that by two different patterns, three packs of six or two packs of nine). 21, 24 (again by two paths), 26, 27, 29, 30 … it looks very much like we’re running out of numbers to buy, and some experimentation finds that 43 is the biggest number of McNuggets which can’t be bought. At least, we can find the formulas for 44, 45, 46, 47, 48, and 49, and obviously, any number above that you can get by buying enough six-packs on top of whatever one of those is.

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