Is it possible to find such a $n\in \mathbb {Z^{+}}$, for given value of $\lambda;$ $\frac {2^{10+\lambda+n}-2^{10+\lambda}-144759}{3^{10}}<349525$

Question

I'm trying to solve a mathematical problem. I expressed the point where I was stuck with the modular arithmetic. Here is my problem;

Is it possible to find such a $n\in \mathbb {Z^{+}}$, for given value of $\lambda,$

$$\frac {2^{10+\lambda+n}-2^{10+\lambda}-144759}{3^{10}}<349525$$

$$2^{10+\lambda+n}-2^{10+\lambda}-144759 \equiv 0 \pmod {3^{10}}$$

Where, $\lambda =1,2,3,...,39366.$

Is there a mathematical way to solve this problem?..I don't know. But, I tried only random value of $\lambda$ and I used discrete logarithm calculator. After a few unfortunate attempts, I gave up.

At least, is it possible to say the existence of such a ​​value of $\lambda$ without making a calculation?

https://www.alpertron.com.ar/DILOG.HTM

Answer 1

You can write the first as $$2^n-1\lt \frac {349525\cdot 3^{10}+144759}{2^{10+\lambda}}\\n \lt \log_2\left(349525\cdot 3^{10}+144759\right)-10-\lambda\\n\le24-\lambda $$ so there are very few $n$s to try. If $\lambda$ is at all large there are not any.

Answer 2

\begin{align} 2^{10+\lambda+n} - 2^{10+\lambda} - 144759 &\equiv 0 \pmod{3^{10}} \\ 1024 \cdot 2^{\lambda+n} - 1024 \cdot 2^{\lambda} - 144759 &\equiv 0 \pmod{3^{10}} \\ 1024 \cdot 2^{\lambda}(2^n - 1) &\equiv 144759 \pmod{3^{10}} \\ 2^n - 1 &\equiv 13116 \cdot 2^{-\lambda} \pmod{3^{10}} \\ 2^n &\equiv 1 + 13116 \cdot 2^{-\lambda} \pmod{3^{10}} \\ \end{align}