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Apr 27, 1997, 3:00:00 AM4/27/97

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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

---------------------------------------------------

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.

--

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

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 ?

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.

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 ?

>

>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.

>

>

>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 :

>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

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>,

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.

>

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