tag:blogger.com,1999:blog-2433841880619171855.post5483252636281574730..comments2020-06-03T07:23:59.656+01:00Comments on Pat'sBlog: Unknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2433841880619171855.post-31887900375838710782009-09-02T19:29:30.630+01:002009-09-02T19:29:30.630+01:00Doctorbinary,
thanks, I hope the typo is correct...Doctorbinary, <br />thanks, I hope the typo is corrected.<br />Now, about the 2^n -1, the binary expansion of 2^n-1 will always be a string of ones, so we can prove that any power of 2^n -1 mod five will NOT have a remainder of four..because the sequence of modulii start at 1 (for 2^1-1) and each new value increases the power by one. <br />We know that the powers of 2 (beyond 2^0) proceed with base five modulii of 2, then 4, then 3, then 1.... and then repeats this same cycle.... <br />so if we start at one, we can anticipate the sequence of base five modulii by seeing that they too will loop, 1+2---> 3 <br />3+ 4 ---> 2 <br />2+3----> 0<br />0+1 --->1<br />and we are back to the start... <br />so it is the first value plus the rhythm of modulus of the sequence of powers of two that guarentee they will always be some number other than four.Pat's Bloghttps://www.blogger.com/profile/15234744401613958081noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-30799588569656839402009-09-02T16:25:43.627+01:002009-09-02T16:25:43.627+01:00Re: "Ends in 4s": Nice! I like the algeb...Re: "Ends in 4s": Nice! I like the algebra. But to show that the base 5 representation has EXACTLY n trailing 4s, you also need to show that the low order digit in the base 5 representation of 2^n - 1 is never 4. I think this is true based on some values I tried: 2^n - 1 in base 5 seems to always end in either 0, 1, 2, or 3 (corresponding to when 2^n - 1 in base 10 ends in 5, 1, 7, or 3, respectively). You'll have to prove it though. (Yes, there's a mod 5 thing going on there, but I think you have to go back and prove that 2^n - 1 in decimal always ends in 5, 1, 7, or 3.)<br /><br />Re: Your relatively prime conjecture: I have not thought about it yet.<br /><br />Re: Wolfram Alpha vs my converter. My converter is more limited in that it only does decimal/binary conversions, but it is arbitrary precision and will thus handle bigger/smaller numbers than Wolfram Alpha.<br /><br />BTW, there's a typo in your post: you say 2n instead of 2^n in the sentence that starts "Rick showed that...".Anonymousnoreply@blogger.com