tag:blogger.com,1999:blog-7691251033406320222.post1235761377784917556..comments2024-03-26T04:19:38.862-07:00Comments on kitchen table math, the sequel: from the SMO junior examCatherine Johnsonhttp://www.blogger.com/profile/03347093496361370174noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-7691251033406320222.post-52073597654183718362010-12-03T17:20:25.239-08:002010-12-03T17:20:25.239-08:00How come there's nothing in this question abou...<a href="http://kitchentablemath.blogspot.com/2010/12/lets-not-and-say-we-did-part-2.html" rel="nofollow">How come there's nothing in this question about Iceland?</a> <br /><br />That's what I want to know.Catherine Johnsonhttps://www.blogger.com/profile/03347093496361370174noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-47546583352367325142010-11-21T10:19:31.852-08:002010-11-21T10:19:31.852-08:00If n is prime, then n^k has k+1 divisors.
Sure. ...<i>If n is prime, then n^k has k+1 divisors.</i><br /><br />Sure. The divisors are precisely:<br /><br />1, n, n^2, n^3, n^4, .... , n^k<br /><br />All you need to know, really, is that any divisor of N can be constructed by choosing a subset of the prime factors of N.<br /><br />So, for example, if N = (p^a)(q^b), where p and q are both prime, then you can build any divisor by taking "some p's" (where "some" is any number from 0 to a) and "some q's", (where "some" is any number from 0 to b). So there are (a+1) choices of how many p's to take, and (b+1) choices of how many q's to take, which makes (a+1)(b+1) possibile divisors you can build.Michael Weissnoreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-36212792476602574042010-11-20T19:32:48.423-08:002010-11-20T19:32:48.423-08:00sorry if this seems like elementary playing -- I&#...sorry if this seems like elementary playing -- I've not been exposed to number theory for many years. It's just something that doesn't come up in the sciences a lot. <br /><br />Do the mainland Chinese pay a lot of attention to number theory? <br /><br />Now I had an Indian teacher (Mr M. Ravi) in sec 1, who exposed me to the SMO. Alas, I've not had a math teacher like that since. Sure I've had useful math teachers, but he was a well-published-adult-with-the-curiosity-of-a-child sort of guy. In fact I think he wasn't a regular teacher at ACSI -- he just randomly ambushed me at a school recess one day, sized me up and said I looked like the kind of person who would be curious enough to attend a mysterious session about something mathy at 1 pm on Fridays.le radical galoisienhttps://www.blogger.com/profile/14684821442296479803noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-78535774499762373512010-11-20T19:19:08.432-08:002010-11-20T19:19:08.432-08:00unproven, but probably true (?) statements:
If n ...unproven, but probably true (?) statements:<br /><br />If n is prime, then n^k has k+1 divisors. <br /><br />If n is a square of a prime, then n^k has 2k+1 divisors. <br /><br />If n has two prime factors (e.g. n=6), then n^k has (k+1)^2 divisors. <br /><br />if n is a cube of a prime (e.g. 8), then n^k has 3k+1 divisors. <br /><br />if n has three prime factors, then n^k has (k+1)^3 divisors.<br /><br /><br />trying to work out the trend (not prove!) for the case to any n.<br /><br />The prime factorization of the number of divisors for any whole number is interesting.le radical galoisienhttps://www.blogger.com/profile/14684821442296479803noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-62254899922721750622010-11-20T18:39:34.229-08:002010-11-20T18:39:34.229-08:00For n^k as well.
Length[Divisors[6^7]] = 64 -- th...For n^k as well.<br /><br />Length[Divisors[6^7]] = 64 -- that is (7+1)^2.<br /><br />So this can be generalised for any n, not just n = 10^b.le radical galoisienhttps://www.blogger.com/profile/14684821442296479803noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-91412401755229129242010-11-20T18:31:43.782-08:002010-11-20T18:31:43.782-08:00interesting. that the sequence of squares alternat...interesting. that the sequence of squares alternate from odd and even -- because the square of an odd number is always odd. <br /><br />but it happens to work out that it makes a formula very convenient, in the case of degenerate pairs.le radical galoisienhttps://www.blogger.com/profile/14684821442296479803noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-20493302155337966782010-11-20T18:29:59.036-08:002010-11-20T18:29:59.036-08:00You can /prove/ that of course, but sometimes I fi...You can /prove/ that of course, but sometimes I find that knowing the proof is not quite the same as conceptual mastery. <br /><br />Using mathematica, Length[Divisors[1000]] gives 16 divisors, but there are 25 digits in its product.<br /><br />Length[Divisors[10000]] ==> 25 divisors<br /><br />(I wrote a short script to investigate this for any n.)le radical galoisienhttps://www.blogger.com/profile/14684821442296479803noreply@blogger.comtag:blogger.com,1999:blog-7691251033406320222.post-10876638294858193912010-11-20T18:11:44.073-08:002010-11-20T18:11:44.073-08:0010^4 is 10000.
Perhaps it is cleaner to show peo...10^4 is 10000. <br /><br />Perhaps it is cleaner to show people that the prime factorization of 10^k is<br />2^k 5^k.Anonymousnoreply@blogger.com