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20210917, 17:06  #1 
Aug 2020
79*6581e4;3*2539e3
399_{10} Posts 
Palindrome prime number of digits is always odd (except 11, and this is true in any base)
This newly discovered largest known Palindrome prime was reported by Propper & Batalov:
10^1234567  20342924302 · 10^617278  1 It's shown as having 1234568 decimal digits which I thought was odd. It should be a nice 1234567 decimal digits, or not? I noticed PARI apparently has a rounding problem: Code:
gp > log(10^73323*10^351)/log(10) %2362 = 72.999999999999999999999999999999999998 gp > log(10^75323*10^361)/log(10) %2351 = 75.000000000000000000000000000000000000 
20210917, 17:21  #2 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2^{3}·5·17 Posts 
10^1 has 2 digits, thus 10^(n) has n+1 digits

20210917, 17:32  #3  
"Rashid Naimi"
Oct 2015
Remote to Here/There
3^{2}×239 Posts 
I learned this from ScienceMan. I wonder what ever happned to him.
In Parigp Quote:
Code:
%4 = 1234567 

20210917, 17:41  #4 
"Viliam Furík"
Jul 2018
Martin, Slovakia
680_{10} Posts 

20210917, 17:49  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3×797 Posts 
...but 10^(n)epsilon has n digits
And of course, the only palindrome prime that has an even number of digits (in any base, actually) is "11"! (And even "11" can be composite; then none.) All others are divisible by "11" (in that base). So of course this number has 1234567 decimal digits. The servercalculated number is in error. The path of least resistance that I did suggest to Prof.Caldwell is  just run pfgw od input awk '{print length($2)}' . (His server runs pfgw for verification on most inputs except most arcane ones.) There is no use making an ad hoc "yet another" calculator with ~million digit precision, when one already has Pari/GP and pfgw. 
20210917, 21:37  #6 
"Rashid Naimi"
Oct 2015
Remote to Here/There
100001100111_{2} Posts 
Apologies for spamming the mods. I think I finally understand that in any other base only 11 (in that base) can be an evennumbereddigitscounted prime. Generally any prime p will equal 11 in base p1 and any other palindromes with even digits count will be divisible by p.
No there is no fire, it's just the smoke coming out my ears. 
20210921, 19:31  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3×797 Posts 
I've actually been asked, so I could repost here exactly why every "evenlength" palindrome is divisible by "11" (think decimal, but it is true for any base b).
We could simply say that order of mod(b,b+1) is 2, and residues are 1,+1. (and everything follows) In other words: First, quick Lemma: 10^n = { +1 (mod 11) if n is even  1 (mod 11) if n is odd }. Proof: For even n, 10^2k = 100^k = +1 (mod 11), because 100 = 1 (mod 11). And more generally for any base b, b^2k = (b^2)^k = 1^k = 1 (mod b+1), because b^2 1 = 0 (mod b+1). For odd n, 10^(2k+1) = 100^k * 10 = 1 * (1) = 1 (mod 11). Now this leads to the (known to some from school) divisibilityby11 criterion: You sum up all digits jumping by 2, let's call it A. (In other words all oddpositioned digits.) Then you add up all evenpositioned digits, let's call it B. Now, if AB is divisible by 11, then the whole number is divisible by 11.* We used to have bus tickets (in the USSR), with sixdigit numbers on them, obtained from a machine (that had a rubber belt so that others could see that you really dropped your 5 cents there), and so, we added those digits and called those where A==B lucky tickets. So those were incidentally divisible by 11. So if you do the above summation for a palindrome with even length, it will be surely lucky (because they will be the same set of digits, because of palindromeness). So it will be divisible by 11. ____ * note that every one knows the divisibilityby3 and 9: "the digital root". This is only one step up from that. One step below that in simplicity is divisibilityby2 and 5. Two steps up is divisibilityby7 and 13. (for it, you would split the long number in 3digit groups and sum them up in your head as groups A and B, then the same as with 11. Why? because 1001 is divisible by 7 and 13 (and 11, again; so you can do all three in one weird sweep). 
20210921, 19:46  #8 
Sep 2002
Database er0rr
3,863 Posts 
Mod 11 is used to get the check digit of ISBNs https://en.wikipedia.org/wiki/Intern...r#Check_digits

20210921, 19:53  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3×797 Posts 
Credit (and SIM) card numbers have the similar (not the same) protection from typos and transpositions.

20210921, 20:25  #10  
Feb 2017
Nowhere
2^{3}·7·89 Posts 
Quote:
An alternate approach is as follows: To any integer base b >1, a baseb palindrome N with 2k digits may be written where the d_{j} are the baseb digits. Since the exponents of b in the sum are all odd, the terms in parentheses are all divisible by b + 1. The digit d_{0} is nonzero because it is also the lead digit. For k > 1, b^{2k  1} + 1 > b + 1, so N/(b + 1) > 1. Last fiddled with by Dr Sardonicus on 20210921 at 20:27 Reason: xingif optsy 

20210922, 19:06  #11 
Aug 2020
79*6581e4;3*2539e3
3×7×19 Posts 
The length has been corrected, now I can sleep peacefully.
Thanks for the explanations about the necessity of an odd number of digits. It's always fascinating how much can be deduced about numbers by using modulus. It never played a big role at school or my two math courses at university. 
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