Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone give me an example of Lucas Theorem and how it works? What about for composite modulus?

share|improve this question
1  
Please provide some more context. Are you talking about this? –  Srivatsan Dec 31 '11 at 18:41
    
Yes, I am; I am much better at learning from examples though –  WhatsInAName Dec 31 '11 at 18:45

1 Answer 1

up vote 3 down vote accepted

Let $p=3$, let $m=44=(1122)_3$, and let $n=14=(112)_3$. Then Lucas's Theorem says that $$\binom{44}{14}\equiv\binom{1}{0}\binom{1}{1}\binom{2}{1}\binom{2}{2}\bmod 3,$$ i.e. $$114955808528\equiv 1\cdot1\cdot2\cdot1\equiv 2\bmod 3.$$ This is correct; when we divide $114955808528$ by $3$, we get $38318602842$ with a remainder of $2$.

share|improve this answer
    
So it's taking the digits of m and n in base p and finding a bunch of mini combinatorics? Is there also a way to do this for nonprime mod? What if p is larger than 10, where digits are no longer single digits? –  WhatsInAName Dec 31 '11 at 19:12
1  
@user1123950: For nonprime modulus, you can use the Chinese Remainder Theorem. For example, if you want a computation mod $105=3\cdot5\cdot7$, you just need to do the computation mod 3, then mod 5, then mod 7 (using Lucas's Theorem) and combine the results. But Lucas's Theorem does not tell you the result modulo higher powers of p. If you want to do a computation modulo $75=3\cdot5^2$, you just need to do the computation mod 3 and then mod $5^2$; but there are no shortcuts as nice as Lucas's Theorem for the mod $5^2$ computation. –  Jonas Kibelbek Dec 31 '11 at 22:55
    
@user1123950: Lucas's Theorem works just fine for primes larger than 10; just be sure you understand base $b$ expansions of integers. If you use $p=17$, you will use base 17 expansions, which use as digits $0, 1, 2, \ldots, 14, 15, 16$. –  Jonas Kibelbek Dec 31 '11 at 22:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.