[ November 10, 2022 ] [ Last update November 18, 2022 ]
Smallest multipliers to make a number palindromic
in bases 2 to 12 (b ≠ 10)
by Xinyao Chen (email)
First of all if this topic is new to you familiarise
yourself with it by studying WONplate 36 and WONplate 96
which deals only with the base 10 variant.
Xinyao Chen went further and delved into
the other bases as well. This gives cause to a lot of new material to list
and lots of opportunity to extend the tables.
Of course, in base b the numbers which are multiples of b will have no
palindromic multiples, since all multiples of these numbers end with 0
but not begin with 0, thus not palindrome.
Only consider the X not divisible by the base, since all multiples of X
divisible by the base has trailing zeros, thus cannot be palindromic
since a number cannot have leading zeros.
For bases 2 to 7, the smallest multipliers of the numbers ⩽ 2^12
(which are not multiples of b) are in the worksheets, and for base 8,
I found the smallest palindromic multiples of all numbers ⩽ 10000
(which is 2^12 in decimal) which are not multiples of 8, except three numbers:
6161 , 6771 , 7601 , they are like the numbers 8181, 8991, 9801 in base 10,
and seems to be hard.
You can also consider the concatenations of 61 in base 8 and the sequence
677...771 in base 8 (they should be like the concatenations of 81 in base 10
and the sequence 899...991 in base 10, respectively).
Also, in base b, a palindromic number is a multiple of b^2b+1 (i.e. 81 in base 10)
if and only if its digits sum is a multiple of b^2b+1, this is why in general,
the least multiplier is larger if the number is a multiple of b^2b+1
Also, it appears that the three largest term below b^4: b^42*b^3+2*b^22*b+1
(i.e. 8181 in base 10), b^4b^3b+1 (i.e. 8991 in base 10), b^42*b^2+1
(i.e. 9801 in base 10) have three largest multipliers, and it appears that
the concatenation of b^2b+1 (i.e. 81 in base 10) in base b and numbers of
the form b^nb^(n1)b+1 (i.e. 8999...9991 in base 10) have larger multipliers.
Instead of showing long lists that might be rather boring I asked Xinyao Chen
for his program that recreates these lists on demand. The pari/gp code is
very intricate and uses functions that call other functions. Here it is:
\\ This is the PARI/GP program code of WONplate 223: (just type f(b)
\\ to print the smallest palindromic (in base b) multiple of the
\\ numbers <= 2^12 not divisible by b) e.g. for b = 8 use (01:54) gp > f(8)
a(n,b)={my(d, i, r);
r=vector(#digits(nb^(#digits(n\(b+1),b)),b)+#digits(n\(b+1),b));
n=nb^(#digits(n\(b+1),b)); d=digits(n,b); for(i=1, #d, r[i]=d[i];
r[#r+1i]=d[i]); sum(i=1, #r, b^(#ri)*r[i])}
c(k,b)=for(n=2,2^64,if(a(n,b)%k==0,return(a(n,b))))
d(n)=if(n<10,n+48,n+55)
e(n,b)=for(k=1,length(digits(n,b)),print1(Strchr(d(digits(n,b)[k]))))
f(b)=for(k=1,2^12,if(k%b,e(k,b);print1(",");e(c(k,b),b);print()))
For bases {6}, {8}, {10}, {12} X. Chen found the smallest palindromic multiples
of the harder numbers (all written in corresponding bases).
41 * 1235_{{6}}  →  55555 
61 * 123457_{{8}}  →  7777777 
81 * 12345679_{{10}}  →  999999999 
A1 * 123456789B_{{12}}  →  BBBBBBBBBBB 

122 * 15304_{{6}}  →  2355532 
142 * 1627405_{{8}}  →  257777752 
162 * 172839506_{{10}}  →  27999999972 
182 * 18293A4B607_{{12}}  →  29BBBBBBBBB92 

203 * 14025_{{6}}  →  3255523 
223 * 641753_{{8}}  →  167777761 
243 * 122633744_{{10}}  →  29799999792 
263 * 15A26B38049_{{12}}  →  38BBBBBBBBB83 

4041 * 5133_{{6}}  →  33355333 
6061 * 7542405_{{8}}  →  56707770765 
8081 * 445372913_{{10}}  →  3599058509953 
A0A1 * A04230888_{{12}}  →  84BB00A00BB48 

4141 * 11534515_{{6}}  →  53555455535 
6161 * 2263561677301_{{8}}  →  16477777677777461 
8181 * 1173450678278939_{{10}}  →  9599999998999999959 
A1A1 * 228355B68929AA3ABB501_{{12}}  →  1A6BBBBBBBBBABBBBBBBBB6A1 

4451 * 102355_{{6}}  →  505545505 
6671 * 55243524_{{8}}  →  466747747664 
8891 * 8546948927_{{10}}  →  75990922909957 
AAB1 * 32A411839712_{{12}}  →  2B3BB0B88B0BB3B2 

4551 * 111222333445_{{6}}  →  555555555555555 
6771 * 111222333444555667_{{8}}  →  777777777777777777777 
8991 * 111222333444555666777889_{{10}}  →  999999999999999999999999999 
ABB1 * 111222333444555666777888999AAB_{{12}}  →  BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB 

5401 * 20304051_{{6}}  →  154555555451 
7601 * 203040506071_{{8}}  →  1767777777777671 
9801 * 2030405060708091_{{10}}  →  19899999999999999891 
BA01 * 2030405060708090A0B1_{{12}}  →  1BABBBBBBBBBBBBBBBBBBAB1 
Xinyao Chen has not confirmed that the numbers for A1A1 (base 12),
6771 (base 8), ABB1 (base 12), BA01 (base 12) are the smallest solutions, but
they are genuine solutions, and these four numbers have no small solutions, the
solutions for these four numbers are only conjectured, found by the law of the
equivalent numbers in base 6 and base 10.
