### Solution to the priates puzzle

Thank you to those who participated in the puzzle.

The solution in words:

If a number of pirates need only one shoe and of the others half need no shoes and the other half need two shoes, then on average each pirate needs one shoe. So, the answer is 100 shoes.

Algebraic solution, thanks to Mr. Person:

Let x = the number of one legged pirates, and let y = the number of pirates with two legs. If all y pirates wore shoes, we would need 2 y shoes for this group.

But only half wear shoes, so we only need y shoes for this group. And since we only need x shoes for the one-legged (one shoe per pirate), the total number of shoes

required turns out to be equal to x + y, which is simply the number of pirates.

Also, thanks to S. Elsnick who pointed out that some assumptions needed to be made in solving this problem: all of the one-legged pirates always wear one shoe; there are no pirates without at least one leg; no pirates have three legs; the remaining half of the two-legged pirates always wear two shoes.

Elias.

The solution in words:

If a number of pirates need only one shoe and of the others half need no shoes and the other half need two shoes, then on average each pirate needs one shoe. So, the answer is 100 shoes.

Algebraic solution, thanks to Mr. Person:

Let x = the number of one legged pirates, and let y = the number of pirates with two legs. If all y pirates wore shoes, we would need 2 y shoes for this group.

But only half wear shoes, so we only need y shoes for this group. And since we only need x shoes for the one-legged (one shoe per pirate), the total number of shoes

required turns out to be equal to x + y, which is simply the number of pirates.

Also, thanks to S. Elsnick who pointed out that some assumptions needed to be made in solving this problem: all of the one-legged pirates always wear one shoe; there are no pirates without at least one leg; no pirates have three legs; the remaining half of the two-legged pirates always wear two shoes.

Elias.

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