If we write down the digits of the $33$ rd Fermat number $$F_{33}=2^{2^{33}}+1$$ in base $10$ in reverse order , the resulting number should , considering its magnitude , be composite.
But can we search for a prime factor WITHOUT actually calculating $F_{33}$ ? $F_{33}$ begins with a $9$ , so there is no obvious prime factor of the reversed number. Can I , for example , determine whether the reverse of $F_{33}$ is divisible by $7$ ?
Best Answer
The Fermat number $F_{33}$ can be calculated in around 3 seconds on my machine. Reversing its decimal representation - we can trial divide by around 1 prime per second. With this, we have that:
$$ (7 \cdot 17 \cdot 37 \cdot 101 \cdot 101 \cdot 191 \cdot 257) \; | \; \text{rev}(F_{33}) = \underbrace{ 7986106331 \cdots 3310530369}_{\text{2,585,827,973 digits}} $$
Here is the code that I used to generate the factors - let me know if you have any questions.