Problems & Puzzles: Puzzles

Puzzle 99. Primes p such that R(p) is prime and p^2 = R(R(p)^2) .
Note: R(p) means "the reversible of p".

The least example is 13, because:

p=13 is prime
R(p)=31 is prime &
13^2 = 169 = R(31^2)= R(961)=169

A rather large of this kind of primes is 3011110001 (Digits =10):

p=3011110001 is prime
R(p)=1000111103 is prime &
3011110001^2 = 9066783438122220001 = R(1000111103^2) = R(1000222218343876609) = 9066783438122220001

The property seems to be not entirely trivial, because:

a) not all primes are such that its reversible are primes
b) not all primes p such that its reversible are primes, are such that p^2 = R(R(p)^2)

For example:

p=1231 is prime
R(p)=1321 is prime but
1231^2=1515361 is not equal to R(1321^2)=R(1745041)=1405471

Questions:

a)    Find an example of D digits for D = 20, 30, 40, 50, ?

b)    It seems that all these type of primes in his decimal expression are composed of digits less than 4. If this is true, can you visualize any other rule about these primes?

Solution

Felice Russo has sent (10-11/07/2000) the following primes for question a):

 digits Prime p 20 10000000000002010001 30 100000000000000000000000020011 40 1000000000000000000000000000000000311001 50 10000000000000000000000000000000000000000000010111

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Faride Firoozbakht wrote (March 2004):

I'm sure that if n is a number such that n^2=R(R(n^2)) the following
three statements are true but I don't know how to explain the reasons.

1.All of the digits of n (in decimal expression) are less than 4.

2.If a and b are two arbitrary digits (in decimal expression) of
n then a+b<5.

3.In his decimal expression of n there doesn't exist three
2 beside each other(222 doesn't appear).

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