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[ May 3, 2002 ]
Powers of Palindromes as products of
integers and their reversals.
by Carlos Rivera


Carlos constructed some general formulas relating
palindromes & reversible numbers in a most interesting manner :

(1)n(4)n(4)n * (4)n(4)n(1)n = [ (2)n(5)n(2)n ]2
(144)n * (441)n = [ (252)n ]2
1(0)n4(0)n4 * 4(0)n4(0)n1 = [ 2(0)n5(0)n2 ]2
15(9)n84 * 48(9)n51 = [ 27(9)n72 ]2

And the puzzle is...
to find some other general formulas !

[ May 5, 2002 ]
Jean Claude Rosa [email] sends three other formulas
of which the first two resemble those of Carlos.

144(0)n144 * 441(0)n441 = [ 252(0)n252 ]2
1(0)n5(0)n8(0)n4 * 4(0)n8(0)n5(0)n1 = [ 2(0)n7(0)n7(0)n2 ]2
276(36)n48 * 84(63)n672 = [ 48(36)n384 ]2

[ May 7-8-11, 2002 ]
Jean Claude Rosa finds it a good fun puzzle
and submits more of the same.

158(558)n4 * 4(855)n851 = [ 277(477)n2 ]2
144(0)n480(0)n4 * 4(0)n084(0)n441 = [ 24(0)n292(0)n42 ]2
(1)n0(4)n0(4)n * (4)n0(4)n0(1)n = [ (2)n0(5)n0(2)n ]2
1(45)n44 * 44(54)n1 = [ 25(45)n2 ]2
(1)n(5)n(8)n(4)n * (4)n(8)n(5)n(1)n = [ (2)n(77)n(2)n ]2
148(988)n84 * 48(889)n841 = [ 269(889)n62 ]2
14(558)n544 * 445(855)n41 = [ 254(774)n52 ]2

"et la suivante, très jolie et très simple"

(1584)n * (4851)n = [ (2772)n ]2

[ May 11-12-13, 2002 ]
Despite being thrilled with all his previous discoveries
Jean Claude Rosa took some distance from his equations
only to notice that some generalisations could be made !

First he sampled the next three methods
Number repetition
Digit repetition
Inserting zeros

144 * 441 = 2522
(144)n * (441)n = [(252)n]2
(1)n(4)n(4)n * (4)n(4)n(1)n = [(2)n(5)n(2)n]2
1(0)n4(0)n4 * 4(0)n4(0)n1 = [2(0)n5(0)n2]2

1584 * 4851 = 27722
(1584)n * (4851)n = [(2772)n]2
(1)n(5)n(8)n(4)n * (4)n(8)n(5)n(1)n = [(2)n(7)n(7)n(2)n]2
1(0)n5(0)n8(0)n4 * 4(0)n8(0)n5(0)n1 = [2(0)n7(0)n7(0)n2]2

12544 * 44521 = 236322
(12544)n * (44521)n = [(23632)n]2
(1)n(2)n(5)n(4)n(4)n * (4)n(4)n(5)n(2)n(1)n = [(2)n(3)n(6)n(3)n(2)n]2
1(0)n2(0)n5(0)n4(0)n4 * 4(0)n4(0)n5(0)n2(0)n1 = [2(0)n3(0)n6(0)n3(0)n2]2

From that data he generalized these results

If Q * R = P2   (R is the reverse of Q)
then (Q)n * (R)n = [(P)n]2 is also true
"The proof of this result is very easy", he adds.

Let Q = abc...xyz ; R = zyx...cba ; P = klm...mlk
then the equality Q * R = P2 leads to

[(a)n(b)n(c)n...(x)n(y)n(z)n] * [(z)n(y)n(x)n...(c)n(b)n(a)n] =
[(k)n(l)n(m)n...(m)n(l)n(k)n]2

[a(0)nb(0)nc...x(0)ny(0)nz] * [z(0)ny(0)nx...c(0)nb(0)na] =
[k(0)nl(0)nm...m(0)nl(0)nk]2

Here are the first ten basic Q numbers which can be applied
generally to the three equation methods described above.
(no other Q's up to 1,000,000 - the R's are not listed)

144
1584
12544
14544
14884
15984
27648
137984
159984
409739

Numbers like 10404, 114444 or 144144 are not included
as basic either since these can be deduced from 144 by applying
one or more of the three described methods.
The three methods are applicable to any number (basic or not)
but the order of application is important. Indeed we have

method + method = method + method
method + method = method + method
but
method + method <> method + method

The law of method composition is not commutative !
as the following example shows

Let starting Q number be 1584
First method 11558844 and then method gives 101050508080404
First method 1050804 and then method gives 11005500880044



A000131 Prime Curios! Prime Puzzle
Wikipedia 131 Le nombre 131














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