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Number 81 is Anti-Palindromic

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Patrick De Geest

Apr 27, 1997, 3:00:00 AM4/27/97
to

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...

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 !?

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

Apr 27, 1997, 3:00:00 AM4/27/97
to

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.

--

Kevin Brown

Apr 27, 1997, 3:00:00 AM4/27/97
to

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

>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

Apr 27, 1997, 3:00:00 AM4/27/97
to

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

Apr 30, 1997, 3:00:00 AM4/30/97
to

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>,

>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.
>

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