20110918, 19:50  #12 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×23×97 Posts 
You were too fast to reply :D, that was a mispost of mine, which I edited already, immediately (2:47AM here, almost sleeping). I was affraid of your fast reply :D I saw you also edited my post, a thing I don't like. Please don't change the meaning of the things I said.
Still did not get the part about prime numbers in base 3 ending in 2. So what? Are they even? Last fiddled with by LaurV on 20110918 at 19:57 
20110918, 20:27  #13  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
I see no counter example proof: the addition in base 3 goes 20*11 = 12 20*22 = 12+20 =32 since 20 affects the second place up and beyond the last digit wouldn't change unless maybe negative. 

20110918, 23:16  #14 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 

20110918, 23:23  #15  
Dec 2010
Monticello
2^{4}·107 Posts 
Quote:
2) To keep track of whether a number in an odd base is even or not, you don't have to do as much as you said....to see this, just take my favorite expansion and ask what happens modulo 2. Things will simplify. 3) On prime numbers in base 3 ending in 2: At some time BEFORE midnight, work out what 6x1 looks like in base 3...you'll get the idea.. SM88: You haven't *quite* got a proof yet, but you are close....try a rewrite  start by stating what you are trying to prove. Last fiddled with by Christenson on 20110918 at 23:26 

20110918, 23:46  #16  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
first what numbers are you looking at ? because : 12_{3}20_{3} = (201)20 = 1 = an answer without a ending 2 as should be all numbers in 6x1 below it. 

20110919, 02:12  #17  
Romulan Interpreter
Jun 2011
Thailand
2^{2}·23·97 Posts 
Quote:
At a second thought, reading my post again, in spite of the fact that I was writing it a 3 o'clock in the night, i clearly stated "evenly split". I may not know the English grammar quite well, but I have the sufficient vocabulary (maybe more then an average native speaker), as I was (a good) scrabble player in my youths and had to learn words by heart, even I didn't know their meaning (learned "heavy" words from a list, that is, words with high scrabble score). And NO, you can't "evenly split" 3 into 2+1. 2. You are most probably right at this part, and I am tempted to believe you. I was a crack at this part. Mea culpa. Posting that was silly, and even sillier was me editing it later, without effectively computing it. I am fully aware of the fact that saying this now improves my crank score Still did not figured it out (just woke up), but I promise you that I will take a paper and a pencil and play with it at my coffee. 3. Here we disagree again. There is NO EVEN PRIME NUMBER except 2. No matter what is the base you write the number in, no matter what is the SYMBOL (2, 7, a, w, etc) that ends the number. The OP stated that "every prime number of the form (whatever) is even". False. And false it will be. I never contested the fact that numbers of the form kb1 end in (b1 as digit) when written in base b. Like you said... "I give you a bonus if you find an EVEN prime other then 2, no matter what base you write the number in, odd base, even base, prime base, etc..." Last fiddled with by LaurV on 20110919 at 02:19 

20110919, 02:55  #18 
Dec 2010
Monticello
2^{4}·107 Posts 
Let me try to collect that bonus: My number is 2^43,112,6091. It's even..it MUST be even........I swear it!
I knew you knew there was a problem with 6x1...SM88 is being his incoherent self in trying to prove the property you give..the property you gave is exactly what I was driving at. And I must have lost an even part of your post somewhere...it's now getting late on me! 
20110919, 03:16  #19 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×23×97 Posts 
Now, a number x (see x as a heap of candies, sticks, coins, not as its numerical representation) can be represented in positional notation in any base b by an expression like:
To be an even number, this has to be congruent to 0, modulo 2. Call this class "the parity" of x. So, the parity is 0, the number is even. The parity is 1, number is odd. The "parity" of this "sum" is given by the parity of its terms. When the base b is even, all the terms are even (as being a product of an even number and something else), except the last one, . So, the "parity" of the number is given by this "last digit". Pardon me for "digit", call it figure (I don't like that, I always associate it in my mind with expressions like "a well known figure in society", or "a figure of speech", "the dove is a figure of peace", "let's figure it out", etc, is more confusing). Call it cipher, as in "any of the (Arabic or else) symbols used to represent numbers in positional notation). In our language we have a special word for it, that translates more like "symbol". Now, when the base is odd, the "parity" of each term is given in fact by its coefficient. If is odd, the term is odd. If is even, the term is even. Now, going back to adding digits, as in the "criterion of divisibility by 3 in base 10", where we have to add all digits, here it does not make too much sense to add the digits that are even, does it? They won't change the parity. Is this what you wanted to say? If so, then I learning something new today, and all our arguing was not in vain. I am quite happy about it. Thanks. This was a good coffee this morning.... Edit: Big revelation: you don't even need to "add" them. Counting is enough. Whaaaa, what a revelation! For example in base 3 you only need to count how many of 1 do appear in the number. In base 5 you have to count 1's and 3's altogether... This coffee was for sure a good one. Leaving for my daily job... Last fiddled with by LaurV on 20110919 at 03:28 
20110920, 21:47  #20  
"Daniel Jackson"
May 2011
14285714285714285714
613 Posts 
Actually, 2^431126091 ends in 1 when expressed in base3. All Mersenne primes do (except 3, which is 10_{3}). In base b, where b is prime, primes can end in anything except 0 (with the exception of b). 911 is 2m mod 3^{n} for n<7, where m is a natural number (even or odd):
911 = 1020202_{3} 911 = 1222_{9} 911 mod 27 = K_{27} 911 mod 81 = K_{81} 911 mod 243 = 182 911 mod 729 = 182 911 mod 2187 = 911 Notice the pattern. 911 leaves an even remainder for all powers of 3 less than or equal to 729, but not any higher powers of 3. Quote:
13 202 2626 41104 564355 11333311 150666343 2322662122 33531600616 50254411644 10140043655335 Note the pattern in the last figure: odd, even, even, even, odd, odd, odd, even, even, etc. repeating with a period of 6. This is because of 1/7's period (0.1428571428571428571428571...). Any prime base p will show this kind of pattern, equal in length to the period of 1/p. This means that all powers of 10 end in 1 when expressed in base 3 because 10^{n}1 is always divisible by 9 (3^{2}). Also note that a million is a palindrome in base7 (11333311) and that 1/911 has a period of 14 when expressed in base 7: 0.000243066642360002430666423600024306664236... This means that 7^{7}+1 is divisible by 911. 911 is also a SophieGermain prime, meaning that 2p+1 is prime, in this case 1823. 911 is a palindrome in base5 (12121). Did you know that the difference between a number and it's reverse is always divisible by 9 911119=792=2*2*2*3*3*11 907709=198=2*3*3*11 90357688675309=360459=9*40051=3^{2}*11^{2}*331 Note that I like 911. It's the first 3 digit prime I learned about (U.S. Emergency Phone Number). Before I learned about primes, I actully tried to factor 8675309 (I was only 4 years old at the time) and found it too difficult. I love Tommy Tutone. Did you know that 911#9110119 is prime? Here it is in base36 with letter values: Code:
244 figures (digits) lb4svav3wehd5ohb1zmy25k24vu41mp0m2jicghe4zsnvbaegtakwg8qh8jsvn2nwmn582atkowqhwzaeo6dgkv41wekvj5kj3gwfwhcgdvgbf7cjvy9875wxe3oob0qnxb01ne79u6csddds841n8c2na15zc4vubbof52nzux5dn2xyyef82ygn6nljaccj3t3260yokaik2w6u6hkr0lt4zr4f8b5udaa02y9q2epy687egfz a=10, b=11, c=12, d=13, e=14, f=15, g=16, h=17, i=18, j=19, k=20, l=21, m=22, n=23, o=24, p=25, q=26, r=27, s=28, t=29, u=30, v=31, w=32, x=33, y=34 z=35 Code:
380 digits 32365034185618020010716991513958659767082943236694935729328906136551863378072625698564549664662002553551382345074856987881103473255815956621387223447804841155658305239172609952384690477550897152816216908432126258545416331558137240581301042339607841611309055744941010472200207903859708352679007908322424571268778054926384062108162417863979176373973320401208882140740392521906172351 Last fiddled with by Stargate38 on 20110920 at 21:49 

20110920, 21:53  #21  
Mar 2006
Germany
2^{2}·23·31 Posts 
Quote:


20110921, 03:34  #22 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×23×97 Posts 
The direction of this topic reminds me old times 20 years ago when it was a fashion (at least in my part of the world) to take people names (influential public persons, movie starts, your colleagues, etc), add the ASCII codes of that names, eventually do some other operation (google for Finagle's constant) and "prove" that the respective man/woman, etc. is the reincarnated devil. I think the joke started with some freaks "proving" that Bill Gates is evil. He used to call himself "Bill Gates the Third" after his father and grandfather being also Bill (no idea if this story is true or fabricated) and if you add his ascii codes, voila! The trick is that you can make (almost) anybody's name add to whatever you like, if you give the right rule. I always liked the number 23, I don't know why, since I was quite young, maybe I liked the sound of it, so you take my (real) name, add 5 (2+3) and I am evil too...
So, take "velociraptor", "mythbusters" or other of the given examples, add whatever Finagle's constant you like at the end of them, and voila! one has his primes. This leads to nothing. Just my two pence. Better talk about some math, so I (and anyone else reading) could learn something new... 
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