Problems & Puzzles: Puzzles Puzzle 43. Palprime1*Palprime2 = Palindrome Last week I started wondering if any palprime can be multiplied by another palprime to produce a palindrome. At first it’s very easy to find the pairs of palprimes: 3*3= 9 Unfortunately very soon the car gets stuck Question: *** Jim Howell (10/3/99) amazingly discovered a surprising solution to this puzzle: a unique Palprime2 that when is multuplied for 757, 787, ..., 17471, produces, in each case, a Palindrome, as requested. This is the unexpected palprime number: Paprime2 = 10000000001000100010000000001 He recognizes that probably this is not the least Palprime2 factor for all the Palprimes1's. *** After studying the Howell's solution, is a kind of easy to look for other solutions. My approach (Carlos Rivera, 12/03/99) was this one: Tha Palprime2 found by Howell can be divided in two sections: the nut and the wings: Paprime2 = 10000000001000100010000000001 The nut should be a palindrome (not necesarily prime!) such that when multiplied by Palprime1 produces a Palindrome. The wing are a collection of "zeros" to left and right of the nut, ending by both sides  with a "one", in such a way that all the number (wingnutwing) is Palprime. Inserting zeros enough you can get a palindrome not only when you multiply a certain Palprime1 for this Palprime2 but also for a collection of Palprime1's. Following this approach I got the following other two Palprime2's: Palprime2 = 10000100000100000100001 This two Palprime2's produces a Palindrome when are multiplied by any of the ten (10) Palprime1's of this puzzle The nut needs not to be a palindrome of the type 1000...1...0001, but then probably the Palprime2 gotten only works for a special Palprime1. By example, the nut 5976795 works only for the Palprime1 = 929 Palprime2 = 1000005976795000001 Giovanni Resta wrote (April, 08):
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