How to calculate modulo of high power of 2

Question

I know there are other such topics but I really can't figure how to calculate the following equation: 2^731 mod 645. Obviously I can't use the little theorem of Fermat since 645 is not a prime number and I can't also do the step by step rising of powers(multiplying by 2) since the numbers are still really big. Is there any way to do calculate the result in a normal way (without the enormous numbers) ? Thanks in advance!

Answer 1

$645 = 15\cdot 43\,$ so we can compute $\,2^{\large 731}\!$ mod $15$ and $43,\,$ then combine them (by CRT or lcm).

${\rm mod}\ 15\!:\,\ 2^{\large\color{#c00} 4}\equiv 1\,\Rightarrow\, 2^{\large{731}}\equiv 2^{\large 3}\,$ by $\,731\equiv 3\pmod{\!\color{#c00}4}$

${\rm mod}\ 43\!:\,\ 2^{\large 7}\equiv -1\,\Rightarrow\,2^{\large\color{#c00}{14}}\equiv 1$ so $\,2^{\large 731}\equiv 2^{\large 3}\,$ by $\,731\equiv 3\pmod{\!\color{#c00}{14}}$

So $2^{\large 731}\!-2^{\large 3}$ is divisible by $15,43\,$ so also by their lcm = product $= 645,\,$ i.e. $\,2^{\large 731}\!\equiv 2^{\large 3}\!\pmod{\!645}$

Answer 2

A commenter mentioned Euler's theorem, which could be useful; but another approach presents itself: successive squaring. Recall that if $a \equiv b \mod c$, then $a^2 \equiv b^2 \mod c$. For example, $2^2 \equiv 4 \mod 645$, $4^2 \equiv 16 \mod 645$, $16^2 \equiv 256 \mod 645$, $256^2 \equiv 391 \mod 645$, and $391^2 \equiv 16 \mod 645$. So we know that $2^{32} = 2^{2^5} \equiv 16 \mod 645$. Now we can keep squaring: $2^{64} = 2^{2^6} = (2^{2^5})^2 \equiv 256 \mod 645$, $2^{128} = 2^{2^{7}} \equiv 391 \mod 645$, $2^{256} \equiv 16 \mod 645$, and $2^{512} \equiv 256 \mod 645$.

We can write $2^{731}$ as $2^{512}2^{219} = 2^{512}2^{128}2^{91} = 2^{512}2^{128}2^{64}2^{16}2^{8}2^22^1$. Modulo $645$, this is therefore $256 \cdot 391 \cdot 256 \cdot 391 \cdot 256 \cdot 4 \cdot 2$. Still pretty big, but remember that we already know that $256^2 \equiv 391 \mod 645$ and $391^2 \equiv 16 \mod 645$, so now we have $391 \cdot 256 \cdot 16 \cdot 4 \cdot 2$. $391 \cdot 256 \equiv 121 \mod 645$. $121 \cdot 16 \equiv 1 \mod 645$. $1 \cdot 4 \cdot 2 \equiv 8 \mod 645$. So our final answer is that $2^{731} \equiv 8 \mod 645$.

We can also do variations on this using successive cubing, or other powers; it's sort of a matter of preference.

Answer 3

$645=3\cdot 5\cdot 43$

$2^{731}\equiv -1^{731}\equiv -1 \equiv 2 \equiv (2+3+3) \equiv 8 \pmod {3}$

$\implies \left(2^{731}-8\right) \equiv 0 \pmod 3$

$2^{731}=2.2^{730}=2.4^{365}\equiv 2 \times -1^{365} \equiv -2 \equiv 3 \equiv (3+5) \equiv 8\pmod {5}$

$\implies \left(2^{731}-8\right) \equiv 0 \pmod 5$

$2^{731}=8^.\left(2^{7}\right)^{104}\equiv 8 \times -1^{104} \equiv 8\pmod {43}$

$\implies \left(2^{731}-8\right) \equiv 0 \pmod {43}$

Therefore,

$\left(2^{731}-8\right) \equiv 0 \pmod {LCM(3,5,43)}$

$\implies \left(2^{731}-8\right) \equiv 0 \pmod {645}$

$\implies 2^{731} \equiv 8 \pmod {645}$