Number 81 is Anti-Palindromic

[Patrick De Geest]

What is going on with the number 81 ?

Recently I started to investigate the following
palindromic topic :

Can every integer X be multiplied with another
integer Y to produce a palindrome P ?

X * Y = P

Note :
For obvious reasons integers ending with a zero
are excluded as leading zero's are not allowed.

If X is itself a palindrome evidently there will
always be the solution Y = 1.

I searched all the numbers X < 100 and for
every integer X, I quite rapidly found an Y value.

Here is an overview of the results :

X Y P
-------------------------------------
1 1 1
2 1 2
3 1 3
4 1 4
5 1 5
6 1 6
7 1 7
8 1 8
9 1 9
10 ~
11 1 11
12 21 252
13 38 494
14 18 252
15 35 525
16 17 272
17 16 272
18 14 252
19 9 171
20 ~
21 12 252
22 1 22
23 7 161
24 29 696
25 21 525
26 19 494
27 37 999
28 9 252
29 8 232
30 ~
31 14 434
32 66 2112
33 1 33
34 8 272
35 15 525
36 7 252
37 3 111
38 13 494
39 15 585
40 ~
41 16 656
42 6 252
43 23 989
44 1 44
45 13 585
46 9 414
47 3 141
48 44 2112
49 7 343
50 ~
51 19 969
52 13 676
53 4 212
54 518 27972 **
55 1 55
56 11 616
57 3 171
58 4 232
59 13 767
60 ~
61 442 26962 **
62 7 434
63 4 252
64 33 2112
65 9 585
66 1 66
67 11 737
68 4 272
69 6 414
70 ~
71 845 59995 **
72 88 6336
73 4 292
74 3 222
75 7 525
76 287 21812 **
77 1 77
78 11 858
79 6 474
80 ~

* 81 ?? ?? Y must be > 1.234.567

82 8 656
83 9 747
84 3 252
85 7 595
86 337 28982 **
87 8 696
88 1 88
89 11 979
90 ~
91 11 1001
92 9 828
93 55 5115
94 3 282
95 55 5225
96 22 2112
97 55 5335
98 7 686
99 19 1881
-------------------------------------

But X = 81 came and if there exists an Y value
it must be greater than 1.234.567.
It seems 81 refuses to become palindromic...
Is 81 the obstructionist ?

Has it got to do with dimensions ?
After all 81 is 3^4 and as some of my webpages
about palindromic numbers can testify, palindromes
thrive in flatland 2D and only scarcely show up in
3D and higher dimensional spheres...

http://www.ping.be/~ping6758/index.html
http://www.ping.be/~ping6758/menu2.htm

But 16 also is a fourth power 2^4 and yet it
produces 272 when multiplied with 17.

X Y P
-------------------------------------
2^4=16 17 272
3^4=81 ?? ??
4^4=256 33 8448
5^4=625 8341 5213125
...
-------------------------------------

And fifth powers
-------------------------------------
2^5=32 66 2112
3^5=243 ?? ??
4^5=1024 396 405504
5^5=3125 1685 5265625
...
-------------------------------------

So it looks that only the powers of 3 ( pw > 3 )
refuses to produce a palindrome.

X Y P
-------------------------------------
3^1=3 1 3
3^2=9 1 9
3^3=27 37 999
3^4=81 ?? ??
3^5=243 ?? ??
3^6=729 ?? ??
-------------------------------------

Is this known ? Can someone prove this conjecture ?

Have I not enough patience and exist there
values Y for these X values. Did I abandonned
the search too soon ?
Can someone clear this out for me !?

Thanks in advance.
--
Patrick De Geest
[mailto:Patrick...@ping.be]
---------------------------------------------------
URL : http://www.ping.be/~ping6758/index.html
---------------------------------------------------

[David G Radcliffe]

Patrick De Geest (Patrick...@ping.be) wrote:
: What is going on with the number 81 ?

:
: Recently I started to investigate the following
: palindromic topic :
:
: Can every integer X be multiplied with another
: integer Y to produce a palindrome P ?
:
: X * Y = P
:
: Note :
: For obvious reasons integers ending with a zero
: are excluded as leading zero's are not allowed.

Every positive integer is a factor of a palindrome, unless it is
a multiple of 10.

If X is not divisible by 2 or 5, then 10^p = 1 mod X, where p=phi(X)
is the number of integers between 1 and X which have no common factor
with X. In this case, put Y = (10^p - 1) / X. Then X * Y = 10^p - 1
is a palindrome.

If X is even, then write X = 2^n * W, where W is odd. Construct a
palindrome Q = 10^n * R + 2^n, where R is the digit reversal of 2^n.
Let p=phi(W). Then 10^(2pn) = 1 mod W, so
S := 1 + 10^(2pn) + 10^(4pn) + ... + 10^(2(W-1)pn) = 0 mod W.
Let P := Q*S. Then P is a palindrome which is divisible by X.

A similar construction can be used when X is a multiple of 5.

Example: If X = 81, then p = 54, and Y = (10^54 - 1)/81.
81 * 12345679012345679012345679012345679012345679012345679 =
999999999999999999999999999999999999999999999999999999.

Example: If X = 112, then X = 2^4 * 7, Q = 610016, and p = 6.
Then P = 610016 0..0 610016 0..0 610016 0..0 610016 0..0 610016 0..0 610016.
The 610016's are placed to ensure that P is divisible by 7, while 610016
itself is divisibly by 16. Therefore P is a multiple of 112.

--
David Radcliffe radc...@alpha2.csd.uwm.edu

[Kevin Brown]

Patrick De Geest (Patrick...@ping.be) wrote:
>: Can every integer X be multiplied with another
>: integer Y to produce a palindrome P ?

On 27 Apr 1997 radc...@alpha2.csd.uwm.edu (David G Radcliffe) wrote:
>Every positive integer is a factor of a palindrome, unless it is

>a multiple of 10...

>
>Example: If X = 81, then p = 54, and Y = (10^54 - 1)/81.
>81 * 12345679012345679012345679012345679012345679012345679 =
> 999999999999999999999999999999999999999999999999999999.

We might also note that phi(3^k) is always even, so we
have phi(3^k) = 2m and (10^2m - 1) = (10^m - 1)(10^m + 1).
Clearly the second term in the factorization is not divisible
by 3, so 3^k must divide 10^m - 1. Similarly, m is always
divisible by 3, and we can show that 10^(m/3) - 1 is divisible
by 3^k, as can be seen from the fact that

k-1
10^(3^k) - 1 = (10-1) PROD [ x^(2*3^j) + x^(3^j) + 1 ]
j=0

The term (10-1) is divisible by 2 powers of 3, and each term
inside the product is divisible by exactly 1 power of 3.
This shows that 3^k not only divides 10^phi(3^k) - 1 (by
Euler's theorem), it also divides 10^(phi(3^k)/6) - 1. For
example, we have

81 * 12345679 = 10^9 - 1 = 999999999

243 * 4115226337448559670781893 = 10^27 - 1

and so on.

[Ilias Kastanas]

In article <336346...@ping.be>,

Patrick De Geest <Patrick...@ping.be> wrote:
>What is going on with the number 81 ?
>
>Recently I started to investigate the following
>palindromic topic :
>

>Can every integer X be multiplied with another
>integer Y to produce a palindrome P ?
>

> X * Y = P
>
>Note :
>For obvious reasons integers ending with a zero
>are excluded as leading zero's are not allowed.
>

>If X is itself a palindrome evidently there will
>always be the solution Y = 1.
>
>I searched all the numbers X < 100 and for
>every integer X, I quite rapidly found an Y value.
>
>Here is an overview of the results :

>X Y P
>-------------------------------------

>54 518 27972 **

Eh, I don't know; no; yes; and yes!

81 * 12345679 = 999999999

That is, 10^9 = k*3^4 + 1. It follows easily that 3^5 divides
10^27 - 1, 3^6 divides 10^81 - 1, etc.

Ilias

[b...@math.umd.edu]

Don't you kjnow anything? You have not even looked at the proper integer
field. You should always use the most fluid and endomorphic-implantitude
transformation to find these integers. Needless to say, the computation of
permutations alone is nearly impossible.

]In article <5k00up$s...@uwm.edu>,

radc...@alpha2.csd.uwm.edu (David G Radcliffe) wrote:

>Patrick De Geest (Patrick...@ping.be) wrote:
>: What is going on with the number 81 ?
>:
>: Recently I started to investigate the following
>: palindromic topic :
>:
>: Can every integer X be multiplied with another
>: integer Y to produce a palindrome P ?
>:
>: X * Y = P
>:
>: Note :
>: For obvious reasons integers ending with a zero
>: are excluded as leading zero's are not allowed.
>

>Every positive integer is a factor of a palindrome, unless it is

>a multiple of 10.
>
>If X is not divisible by 2 or 5, then 10^p = 1 mod X, where p=phi(X)
>is the number of integers between 1 and X which have no common factor
>with X. In this case, put Y = (10^p - 1) / X. Then X * Y = 10^p - 1
>is a palindrome.
>
>If X is even, then write X = 2^n * W, where W is odd. Construct a
>palindrome Q = 10^n * R + 2^n, where R is the digit reversal of 2^n.
>Let p=phi(W). Then 10^(2pn) = 1 mod W, so
>S := 1 + 10^(2pn) + 10^(4pn) + ... + 10^(2(W-1)pn) = 0 mod W.
>Let P := Q*S. Then P is a palindrome which is divisible by X.
>
>A similar construction can be used when X is a multiple of 5.
>

>Example: If X = 81, then p = 54, and Y = (10^54 - 1)/81.
>81 * 12345679012345679012345679012345679012345679012345679 =
> 999999999999999999999999999999999999999999999999999999.
>