b + s + p = 100

10b + 5s + 0.5p = 100

And it’s “impossible” to solve that directly using just symbolic manipulation. But as you’ve figured out, you can find an answer with some guess-and-check, or by spending some time with Diophantus.

]]>“A certain buyer said: ‘I want to buy 100 pigs with 100 denarii in such a way that a mature boar is bought for 10 denarii; a sow for five denarii; and two small female pigs for one denarius.’ Let him say, he who knows, How many boars, sows, and small female pigs should there be so that there are neither too many nor too few of either [pigs or denarii]?”

This is in the same vein as the bicycle/tricycle problem, but I found the library’s method more useful on this one than setting up a system of equations.

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