WON plate131 | World!OfNumbers [ May 3, 2002 ] Powers of Palindromes as products of integers and their reversals 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 formulasof 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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