

Title: Binary Implications and Binomial Parity Post by william wu on Oct 29^{th}, 2003, 7:35pm Let r and s be natural numbers. Denote the binary expansion of r by r_{1}r_{2}...r_{n}. Similarly, denote the binary expansion of s by s_{1}s_{2}...s_{n}. Assume these binary expansions are both of length n. Let the relation r [smiley=rrightarrow.gif] s mean that the following sentence is true: Thus for example, for binary strings 1010 and 1110, it follows that 1010 [smiley=rrightarrow.gif] 1110. Prove that r [smiley=rrightarrow.gif] s IF AND ONLY IF C(s,r) is odd where C(s,r) is the binomial coefficient: (s!)/(r!(sr)!) Note 1: An amazing result from the 1900s! It was passed to me via a colleague (Jon Yard). Does anyone know how it is proven (I don't), or what the theorem's name is if it has a name? If anyone knows the proof and feels it is too hard for recreational math, please say so. Note 2: This result apparently has a little to do with Hilbert's 10th problem. 

Title: Re: Binary Implications and Binomial Parity Post by Rezyk on Oct 29^{th}, 2003, 10:44pm Tool #1: [hide]The number of times 2 divides X! is equal to X minus the number of 1s in the binary representation of X.[/hide] (proof omitted) Tool #2: [hide]When adding 2 numbers in binary, iff there is no carry anywhere, the number of "1" digits is conserved.[/hide] (proof omitted) C(s, r) is odd [bigleftrightarrow] [hide]s!/(r!(sr)!) is not divisible by 2[/hide] [bigleftrightarrow] [hide]number of times 2 divides s! = (number of times 2 divides r!)+(number of times 2 divides (sr)!)[/hide] [bigleftrightarrow] [hide]s  (number of 1s in binary repr. of s) = r  (number of 1s in binary repr. of r) + sr  (number of 1s in binary repr. of sr)[/hide] [bigleftrightarrow] [hide]number of 1s in binary repr. of s = (number of 1s in binary repr. of r) + (number of 1s in binary repr. of sr)[/hide] [bigleftrightarrow] [hide]there is no carry anywhere in adding r to (sr) in binary[/hide] [bigleftrightarrow] r [smiley=rrightarrow.gif] s 

Title: Re: Binary Implications and Binomial Parity Post by Barukh on Oct 30^{th}, 2003, 11:05am Awesome solution, Rezyk! And really elementary. Let me add something for the completeness of the derivation: Tool #1  proof: [smiley=blacksquare.gif] [hide]Denote F(n) the maximal power of 2 dividing n!, P(n)  number of 1's in the binary representation (BR) of n. We want to show that F(n) + P(n) = n. Use induction by n: 1. F(1) + P(1) = 1 + 0 = 1. 2. Let the rightmost 1 in BR of (n+1) be at position j. Then, the jth position in BR of n was 0, but positions 1, ..., j1 were all 1's. So, P(n+1) = P(n)  (j2). Also, 2^{j1} divides n+1, so F(n+1) = F(n) + (j1). Therefore, F(n+1) + P(n+1) = n+1.[/hide] [smiley=blacksquare.gif] 

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