Reversed binary representation of primes

Question

I discovered a very interesting fact that if a prime is represented in a binary form , and if that binary is rewritten in the reverse order and converted back to base 10 decimal and if the resulting number is not divisible by 5 or 7 then it is a prime or a square number or a product of different primes

For example:

$(29)_{10} = (11101)_{2}$

Now by rewriting $(11101)_{2}$ in reverse order we get,

$(10111)_{2} = (23)_{10}$

If we perform the same operations other numbers,we get,

$(3)_{10} \rightarrow (3)_{10}$

$(11)_{10} \rightarrow (13)_{10}$

$(17)_{10} \rightarrow (17)_{10}$

$(71)_{10} \rightarrow (113)_{10}$ ($113$ is a prime)

$(149)_{10} \rightarrow (169)_{10}$ ($169 = 13\times 13$)

$(151)_{10} \rightarrow (233)_{10}$ ($233$ is a prime)

So I was wondering if there are any proofs for this property of primes. Can you tell me whether it is true for all primes?

If you could provide any proofs or counter-examples it would be of great help.

Answer 1

Counter examples:

$(607)_{10} \rightarrow (1001)_{10} = 7 \cdot 11\cdot 13$

$(5689)_{10} \rightarrow(5005)_{10} =5 \cdot 7 \cdot 11 \cdot 13$

$(135559)_{10} \rightarrow(230945)_{10} =5 \cdot 11 \cdot13 \cdot 17 \cdot 19$

Note: My answer has counter examples to an earlier version of the question.

Answer 2

More counterexamples: $139$ is 10001011, reversed it's 11010001 which is $209 = 11 \times 19$, not divisible by $5$ or by $7$, not a square. $191$ is 10111111, reversed that's 11111101 which is $253 = 2^8 - 3 = 11 \times 23$.

You've been fooled by what humans call Richard K. Guy's law of small numbers. Mwahahahaha!