Find the reminder of $2^{62}$ with 85

Question

As title, how do you find the reminder of $2^{62}$ with 85?

Since (85,2) = 1, from FLT I know that $2^{84} \equiv 1 \mod(85)$, but after this I don't know how to proceed because 62 < 85 and I don't know how to use this theorem.

Answer 1

$2^{84}$ is actually equal to $16$ mod $85$.

$85$ is not prime, so you need Euler's Theorem, not Fermat's. This says that if $n$ and $a$ are co-prime, then $a^{\varphi(n)}\equiv 1$ mod $n$, where $\varphi(n)$ is the number of integers between $1$ and $n-1$ that are relatively prime to $n$. Note that if $n$ is prime, then $\varphi(n)=n-1$, which gives us Fermat's Little Theorem.

If $n=pq$ is a product of two primes, then $\varphi(n)=(p-1)(q-1)$ (this is the basis of RSA cryptography). So we have $\varphi(85)=4\cdot 16=64$, which means that $2^{64}\equiv 1$ mod $85$.

So $2^{62}\equiv 1/4$ mod $85$. And $4\cdot 64\equiv 1$ mod $85$, so $1/4\equiv 64$ mod $85$.

Answer 2

$85$ is not prime, FLT is not applicable directly

Use http://mathworld.wolfram.com/CarmichaelFunction.html

$\lambda(85)=16$

$\implies2^{64}\equiv1^4\pmod{85}$

$\implies2^{62}\equiv2^{-2}\equiv-21\equiv-21+85$

as $-21\cdot4\equiv1\pmod{85}$

Answer 3

As noticed we can't apply FLT, to solve by elementary methods note that

$$2^{8}=256 \equiv 1 \pmod{85}$$

then

$$2^{62} \equiv 2^{7\cdot 8}\cdot 2^{6}\equiv 2^{6}\equiv 64 \equiv -21 \pmod{85}$$

Answer 4

$\begin{align}{\rm Note}\qquad\ \ \ \ \ 85 &= 4^{\large 3}\!+4^{\large 2}\!+4+1\\ \Rightarrow\ \ (4\!-\!1)85 &= 4^{\large 4}-1\ \Rightarrow\ \color{#c00}{4^{\large 4}\equiv 1}\!\!\!\pmod{\!85}\end{align}$

So $\bmod 85\!:\,\ 2^{\large 62}\equiv 4^{\large 31}\equiv 4^{\large 3}(\color{#c00}{4^{\large 4}})^{\large 7}\equiv 4^{\large 3} \color{#c00}1^{\large 7}\equiv 64$

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$\begin{align}{\bf Or}\qquad\qquad 85&= (16\!+\!1)\cdot 5\\ \Rightarrow\ \ \ 85\cdot 3 &= (16\!+\!1)(16\!-\!1) = 16^{\large 2}-1\,\Rightarrow\,\color{#c00}{16^2\equiv 1}\pmod{\!85} \end{align}$