A Question

[S-496]

I'm an English major who decided to take a history of math class this semester, since I enjoy studying history. At the end of class today, my professor asked us this:

"Are there infinitely many n so that (2n)!/((n!)((2n-n)!)) is not an integer multiple of 3, 5, or 7?"

I'm rather unfamiliar with factorials, so I thought I'd ask for advice. We're not required to solve this problem--it is, after all, a history course that just has to do with math. Nonetheless, I'd like to see how it's done, or at least some observations on how I could solve it. I'm not entirely sure whether this belongs in Advanced Math, this question coming from a history course rather than a math one, but I decided to play it safe and post it here.

 

[lookagain]

I'm an English major who decided to take a history of math class this semester, since I enjoy studying history. At the end of class today, my professor asked us this:

"Are there infinitely many n so that (2n)!/((n!)((2n-n)!)) is not an integer multiple of 3, 5, or 7?"

As of the posting of this link, it was unsolved:

http://centraledesmaths.uregina.ca/QQ/d ... doan1.html

I don't understand why your professor would ask you about an unsolved problem, unless he/she
wanted you to search for information about it on the Internet (not including a math forum
such as this).

 

[S-496]

From the content of the in-class discussion that immediately preceded the question, I suspect that he merely wants observations on the problem, not a solution itself.

 

[daon]

It simplifies to (2n-1)!/(n)!*1/(n-1)! = (2n-1)(2n-2)...(n+1)/ [(n-1)(n-2)...(2)(1)]

= (2n-1)/(n-1) * (2n-2)/(n-2) * (2n-3)/(n-3) * (2n-4)/(n-4) * ... * (n+3)/3 * (n+2)/2 * (n+1)/1

= \(\displaystyle \Big\Pi_{i=1}^{n-1} \frac{2n-i}{n-i}\)

Some simplification may be done. One observation is that the last three factors are an integer. Of three consecutive integers (n+1)(n+2)(n+3), it must be divisible by 2 and 3. Hence we may write it as:

= \(\displaystyle 6k \Big \Pi_{i=1}^{n-4} \frac{2n-i}{n-i}\)

There is apparent cancellation as well, for terms such as the numerator 2n-4 when i=4, and the denominator n-2 when i=2.

[6k*(2n-2)/(n-1) * (2n-4)/(n-2) * (2n-6)/(n-3) * ... ends at some (2n-2r)/(n-r)] * [(2n-1)/(n-r-1)*(2n-3)/(n-r-2)*(2n-5)/(n-r-3)... ends at either (n+4/4) or (n+5)/5] = 6k*2^(n-r) * [second factor]

r will depend on the parity of n. If n is even, r=(n/2-1).

 

[mmm4444bot]

I suspect that [the instructor] merely wants observations

Or, perhaps, he wants you to "put yourself" into the appropriate historical period and to try think from that perspective. In other words, how did the knowledge of the day shape the development of mathematical thought moving forward -- using this exercise as an example.