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 (1011/07/2000) the following primes for question a):
digits 
Prime p 
20 
10000000000002010001 
30 
100000000000000000000000020011 
40 
1000000000000000000000000000000000311001 
50 
10000000000000000000000000000000000000000000010111 
***
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).
***
