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How can $3$ friends pay $5$ cents each to enter some place, when they only have coins of $10$, $15$ and $20$ cents (unlimited of them)?

I started by noticing that each of them must pay an odd quantity and only giving a coin of $15$ to another person changes the parity of both people(the one who gives and the one who receives. So if every one starts at paying an even quantity ($0$), some changes must be done to ensure every one of them has payed an odd quantity and then paying with coins of $10$ or $20$ until you have payed all of it, but this is imposible as you cannot go from $000_2$ to $111_2$ changing two digits at a time.

So here is the problem, my conclusion is that this is not possible, but the problem is in a book of problems, which the logical thing is that it has a valid answer.

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    Let the people be labeled $A,B,C$. Have person $B$ give $A$ a 15cent piece and $A$ gives $B$ a 10cent piece. Similarly $C$ gives $A$ a 15cent piece and $A$ gives $C$ a 40cent piece. Now, $A$ pays the entry fee of 15cents to the doorman. All of $A,B,C$ each have their net worth decreased by exactly 5cents. – 2017-02-26
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    @MrYouMath What should you put as title? – 2017-02-26
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    @JMoravitz submit that as an answer please – 2017-02-26
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    @JMoravitz also you have a typo "$A$ gives $C$ a$ \40$ should be 10 cent piece – 2017-02-26

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