[ *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)_{n}4(0)_{n}4 * 4(0)_{n}4(0)_{n}1 = [ 2(0)_{n}5(0)_{n}2 ]^{2}

15(9)_{n}84 * 48(9)_{n}51 = [ 27(9)_{n}72 ]^{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)_{n}144 * 441(0)_{n}441 = [ 252(0)_{n}252 ]^{2}

1(0)_{n}5(0)_{n}8(0)_{n}4 * 4(0)_{n}8(0)_{n}5(0)_{n}1 = [ 2(0)_{n}7(0)_{n}7(0)_{n}2 ]^{2}

276(36)_{n}48 * 84(63)_{n}672 = [ 48(36)_{n}384 ]^{2}

[ *May 7-8-11, 2002* ]

Jean Claude Rosa finds it a good fun puzzle

and submits more of the same.

158(558)_{n}4 * 4(855)_{n}851 = [ 277(477)_{n}2 ]^{2}

144(0)_{n}480(0)_{n}4 * 4(0)_{n}084(0)_{n}441 = [ 24(0)_{n}292(0)_{n}42 ]^{2}

(1)_{n}0(4)_{n}0(4)_{n} * (4)_{n}0(4)_{n}0(1)_{n} = [ (2)_{n}0(5)_{n}0(2)_{n} ]^{2}

1(45)_{n}44 * 44(54)_{n}1 = [ 25(45)_{n}2 ]^{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)_{n}84 * 48(889)_{n}841 = [ 269(889)_{n}62 ]^{2}

14(558)_{n}544 * 445(855)_{n}41 = [ 254(774)_{n}52 ]^{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 = 252^{2}

(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)_{n}4(0)_{n}4 * 4(0)_{n}4(0)_{n}1 = [2(0)_{n}5(0)_{n}2]^{2}

1584 * 4851 = 2772^{2}

(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)_{n}5(0)_{n}8(0)_{n}4 * 4(0)_{n}8(0)_{n}5(0)_{n}1 = [2(0)_{n}7(0)_{n}7(0)_{n}2]^{2}

12544 * 44521 = 23632^{2}

(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)_{n}2(0)_{n}5(0)_{n}4(0)_{n}4 * 4(0)_{n}4(0)_{n}5(0)_{n}2(0)_{n}1 = [2(0)_{n}3(0)_{n}6(0)_{n}3(0)_{n}2]^{2}

From that data he generalized these results

If Q * R = P^{2} (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 = P^{2} 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)_{n}b(0)_{n}c...x(0)_{n}y(0)_{n}z] * [z(0)_{n}y(0)_{n}x...c(0)_{n}b(0)_{n}a] =

[k(0)_{n}l(0)_{n}m...m(0)_{n}l(0)_{n}k]^{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