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Possible Duplicate:
Lucas Theorem but without prime numbers

This question mentions a strategy for computing C(n, k) modulo a composite number, but leaves out the details. The use of the Chinese Remainder Theorem is standard, but I don't follow the suggestion for how to use the extended version of Lucas' Theorem to compute C(n, k) mod p^q.

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    You saw the above answer, but did *not* link to it in your question and did not elaborate what you did not understand. This makes it impossible to give you a better answer that is not already in the other thread.2012-11-28

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