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
5*11= 55
7*11=77
11*3=33
101*3=303
131*3=393
151*11=1661
…

Unfortunately very soon the car gets stuck

Question:
Can you find the corresponding palprime2 for the following palprime1’s: {757, 787, 797, 919, 929 13931, 15551, 16561, 16661, 17471}?

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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.

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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 (wing-nut-wing) 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
Palprime2 = 100000100000010000001000001

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):

I've checked if by chance there was a
Palprime2 smaller than your P=10000100000100000100001
for any single number in S={757,787,797,919,929,13931,15551,
16561,16661,17471} or for the set as a whole.

The answer is negative, so P is the minimum for every
palprime in S (and thus also for all of them simultaneously).

This also means that 1000005976795000001 unfortunately
it is not a Palprime2 for 929, as you stated.

Indeed the product is -almost- a palindrome:
929005552442555000929, but it has a zero too much
in the right half.

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