Under mod $10^k$, period of $2^n$ and solution of $2^n=2$

Question

On one digit, $2^n$ has a period of 4: 2, 4, 8, 6. So $n=5$.

On two, the period is 20: 2, 4, 8, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 4. But 2 is by-passed. So there is no solution .

In a general case, for $k$ digits, how long is $2^n(mod 10^k)$’s period, and will $2$ be passed?

Answer 1

For $k$ digits the period is $4\cdot 5^{k-1}$ and $k-1$ numbers starting with $2$ are bypassed. The easy way to see that $k-1$ numbers must be bypassed is that all the numbers in the cycle of $k$ digits must be multiples of $2^k$. The first $k-1$ numbers are not multiples of $2^k$ The cycles includes all multiples of $2^k$ that do not end in $0$.

Answer 2

$2^n = 2 + m*10^k$

$2^{n-1} = 1 + 5m*10^{k-1}$

RHS is odd unless $k=1$.

So this is only possible for $n = 1$ and $2^1 \equiv 2 \mod 10^k$ (which isn't cycling) or if $k =1$.

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Furthermore $4|10^k$ for $k>1$ so $4\not \mid 10^k*m + 2$. But $4|2^n$ for all $n > 1$.