remainder of $(2^{87}-1) \div 89$

Question

What will be the remainder when $2^{87} -1$ is divided by $89$?

I tried it solving by Euler's remainder theorem by separating terms:

$$ \frac {2^{87}}{89} - \frac{1}{89}$$

$\phi (89) =88 $

remainder $\dfrac{{87}}{88} = 87;$

this led me to the point from where I started.

Answer 1

So $$2^{88}\equiv 1\pmod {89}$$ and $(2,89)=1$ so $$2(2^{87}-1)=2^{88}-2\equiv 1-2=-1 \pmod {89}$$ so $$2^{87}-1\equiv -2^{-1}\pmod {89},$$ but $2\times 45=90\equiv 1 \pmod {89} and so$ it should be 44.

Answer 2

$2^{88}\equiv1\pmod{89}\implies45\cdot2^{88}\equiv45\pmod{89}\implies90\cdot2^{87}\equiv45\pmod{89}$ Continue from here.

Answer 3

$89$ is prime, so by Fermat's little theorem $2^{88}\equiv 1\pmod {89},$ so $2^{87}\equiv 2^{-1} \pmod {89}$.

Now $45\times 2 = 90 \equiv 1 \pmod {89},$ so this means $2^{87}\equiv 45 \pmod {89},$ so $2^{87}-1\equiv44 \pmod {89}$.