This is a generic generating functions problem, and you are left with 1/(x-1)(x^5-1)....(x^50-1) however now I want to find the coefficient of the x^100 term of this expression... how can I do this or do I have to use partial fractions which will take ages...
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I guess you'd have to use complex numbers... but this will take forever... – grigori Nov 23 '16 at 10:08
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I am guessing your "..." may have $1,5,10,25,50$ cent coins, though it rather depends on which dollar you are assuming. – Henry Nov 23 '16 at 10:12
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Multiplying out $(1+x+x^2+\cdots+x^{100})(1+x^5+\cdots+x^{100})\cdots(1+x^{50}+x^{100})$ is not that difficult with a CAS or even a spreadsheet – Henry Nov 23 '16 at 10:14
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yeh that's right, which dollar you are assuming??? – grigori Nov 23 '16 at 10:15
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And yes that's easier than complex numbers ect...just not very elegant – grigori Nov 23 '16 at 10:15
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https://en.wikipedia.org/wiki/Dollar#Economies_that_use_a_dollar – Henry Nov 23 '16 at 10:16
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Zimbabwe dollar – grigori Nov 23 '16 at 10:17
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This question is essentially a duplicate of https://math.stackexchange.com/questions/15521/making-change-for-a-dollar-and-other-number-partitioning-problems – Hans Hüttel Feb 08 '20 at 08:09
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See http://www.maa.org/frank-morgans-math-chat-293-ways-to-make-change-for-a-dollar. This site shows how the answer is obtained with just some intelligent counting. Despite the title of the site Frank Morgan suggests $292$ ways, not counting a dollar coin.
Oscar Lanzi
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