The answer to our problem ( 293) is the coefficient of x 100 in the reciprocal of the following: ( 1 − x ) ( 1 − x 5 ) ( 1 − x 10 ) ( 1 − x 25 ) ( 1 − x 50 ) ( 1 − x 100 )

Question

The answer to our problem (  293) is the coefficient of       x          100       in the reciprocal of the following:  (  1  −  x  )  (  1  −      x    5    )  (  1  −      x          10        )  (  1  −      x          25        )  (  1  −      x          50        )  (  1  −      x          100        )

Harper Brewer · Accepted Answer

Note, if you are multiplying out to find the coefficient of       x          100      , it would be easier not to go to the reciprocal, which arises from considering an infinite number of terms.You just need to multiply out  (      ∑          j      =      0              100            x    j    )     (      ∑          j      =      0              20            x          5      j        )     (      ∑          j      =      0              10            x          10      j        )     (      ∑          j      =      0              4            x          25      j        )     (  1  +      x          100        )which would amount to enumerating the different ways to make the change (and in fact is the the way we come up with the generating function in the first place).You could potentially do other things, like computing the       100          t      h       derivative at 0, or computing a contour integral of the generating function divided by       x          100      , but I doubt they went that route either.

Felix Cohen · Accepted Answer

You calculate   [      x          100        ]  (  1  −  x      )          100        (  1  −      x    5        )          20        (  1  −      x          10            )          10        (  1  −      x          25            )    4    (  1  −      x          50            )    2    (  1  −      x          100        ), but that calculation seems to be brute force.

The answer to our problem (293) is the coefficient of x^100 in the reciprocal of the following: (1−x)(1−x^5)(1−x^10)(1−x^25)(1−x^50)(1−x^100)

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