tag:blogger.com,1999:blog-4679715190635795973.post2893172918904834636..comments2023-09-06T08:23:30.799-07:00Comments on y of x: Prime Timefakehttp://www.blogger.com/profile/12859709233583742208noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4679715190635795973.post-33345470300720217792009-07-07T02:48:01.140-07:002009-07-07T02:48:01.140-07:00Really clever! I love this and you even got a more...Really clever! I love this and you even got a more general fact. Sorry it took me so long to look this over.fakehttps://www.blogger.com/profile/12859709233583742208noreply@blogger.comtag:blogger.com,1999:blog-4679715190635795973.post-87182439843064739212009-05-22T15:37:32.668-07:002009-05-22T15:37:32.668-07:00I guess this problem has been kicking around in th...I guess this problem has been kicking around in the back of my head long enough it now seems easy (I wish that would happen with my thesis!).<br /><br />Given d+1 integers between 0 and 9, called a_0, ... a_d, let me write A=(a_d a_{d-1} ... a_0) for the number \sum_{i=0}^{d}a_i 10^i.<br /><br />Consider the product obtained by splitting this digit string into (a_d ... a_s)(a_{s-1} ... a_0). So this is a product of a d-s+1 digit number and an s digit number (take s >= 1).<br /><br />Now A=(a_d ... a_0) = 10^s(a_d ... a_s) + (a_{s-1} ... a_0) is at least as big as 10^s(a_d ... a_s). However, since 10^s is strictly bigger than (a_{s-1} ... a_0), A >= (a_d ... a_s)10^s > (a_d ... a_s)(a_{s-1} ... a_0).<br /><br />So any integer is bigger than the product obtained by splitting it's decimal string into two pieces. It seems the general statement, splitting the decimal string into any number of pieces, follows by induction.sumidiothttps://www.blogger.com/profile/14998929191458452400noreply@blogger.comtag:blogger.com,1999:blog-4679715190635795973.post-38309170851076714842009-04-09T19:08:00.000-07:002009-04-09T19:08:00.000-07:00I know what you mean the products always seem to t...I know what you mean the products always seem to turn out too low. To illustrate that and eliminate some possibilities, any number of 7 or more digits cannot have all its primes grouped in single digits. At most a number with seven digits and grouped as single digit primes could be the product of seven sevens, or 7^7, which is 823,543 a six digit number. Look at the progression of 7's<BR/>(7)(7)=49<BR/>(7)(7)(7)=343<BR/>(7)(7)(7)(7)=2401<BR/>7^5= 16,807<BR/>7^6 = 117,649<BR/>7^7 = 823,543<BR/>Each time the exponent increases from n to n+1, there is less real estate in the n+1 digit numbers. n = 3, only 27% of the numbers are even possibilities. When n=4, around 16%; n = 5, 8% and so on...<BR/><BR/>Looking at it more analytically and remembering that if we add one to the floor of log (x) we get the number of digits in x, it is true for any single digit prime p, we must have log(p^n)>n-1. In the previous example p=7 and the inequality is not satisfied for all n>=7. It is a little late for me, so I might have slopped this up, but in general I think for any prime of b digits, log(p^n)>n/b-1. 97 is the highest two digit prime. I wonder for what value of n the inequality will fail? That's all I have tonight.fakehttps://www.blogger.com/profile/12859709233583742208noreply@blogger.comtag:blogger.com,1999:blog-4679715190635795973.post-69667156022883816432009-04-09T09:47:00.000-07:002009-04-09T09:47:00.000-07:00Interesting question! I've been vaguely trying to ...Interesting question! I've been vaguely trying to learn python recently, and thought this would be a good exercise. The loop I wrote wasn't particularly clever, but didn't turn up any results. Neither did the extension to 4 digits. Guess that means it's time to try to prove I didn't miss any :)<BR/><BR/>It seems, to me, pretty reasonable that there aren't any composites in this set. The products you look at never seem to be big enough.sumidiothttps://www.blogger.com/profile/14998929191458452400noreply@blogger.com