I'm examining the behavior of the modulus of powers of 2 and I'm confused on how to prove that these observations are true.

  1. Consider the sequence $(a_n)_{n \in \mathbb{N}}$ defined by the last two digits of powers of 2. Prove that $a_n + a_{n+100} = 100$ for every $n \geq 2$.

  2. Prove that the sequence defined by the last three digits of the powers of 2 (starting with 008) is periodic with period 100.

Using Python, I've calculated these values and graphed them, and the behavior is clearly periodic. For example, for question (1), $a_3 + a_{103} \neq 100$, so I'm unsure if there's something broken here. And for (2), I can show through Python code that the function repeats, but I'm not sure how I would show this for all $n \in \mathbb{N}$, I'm assuming through induction.

  • 1
    $\begingroup$ Part 1 was probably meant to be $$a_n + a_{n+10} = 100$$ for every $n \geq 2,$ which is true and follows from $2^{10} \equiv -1 \pmod{25}$ $\endgroup$ Oct 4 '19 at 2:49
  • 1
    $\begingroup$ I'm pretty sure you can do the first one by induction (haven't tried to verify it though), and the second one is a simple period finding method. You assume a period $t$, and then try to prove that there exists a real value for $t$ such that $2^k \equiv 2^{k+t} \pmod{1000}$ holds for all $k$ $\endgroup$
    – Hiten
    Oct 4 '19 at 2:50
  • $\begingroup$ Thought I was going crazy - @BrianMoehring that works out. And I will try it out right now, thanks for the help! $\endgroup$
    – beflyguy
    Oct 4 '19 at 2:53

For part 2:

Hint: The question is essentially asking for the smallest positive integer $n$ such that $2^n \equiv 1 \pmod {125}$.

We know from corrected part 1 that $ 2^ {10} \equiv -1 \pmod{25}$. Now verify that this is the smallest $n$ for mod 25.

Hint: Hence conclude that $2^{100} \equiv 1 \pmod{125}$ is the smallest $n$.

So, the period is 100, starting from $a_3$.

For part 1, Brian's observation works directly.
Alternatively, you can easily modify the above.

(Sorry, my previous version thought you were still summing the last 3 digits in part 2)


For 2, the sequence must be periodic because there are only finitely many residues $\bmod 1000$. Once you hit one that you have hit before you have found the period. Once you get to $2^3$ all the powers $\bmod 1000$ must be multiples of $8$ so the cycle is no longer than $125$. In fact it is $100$ as you say because the powers also cannot have a multiple of $5$ in them, which leaves $100$ choices.

More generally, the powers of $2 \bmod 10$ repeat with a cycle of $4-\ 2,4,8,6$. The powers of $2 \bmod 10^n$ repeat with a cycle of $4\cdot 5^{n-1}$ because each of the $n-1$ digit numbers that has a factor of $2^{n-1}$ can be extended in $5$ ways to an $n$ digit one that has a factor of $2^n$.

This means $1$ is incorrect. The powers of $2$ cycle with period $20$, so after $100$ your are five times around. For example, $2^5=32, 2^{105} \equiv 32 \pmod {100}$, so their sum is $64$, not $0$

  • $\begingroup$ Actually, the powers of 2 mod 100 cycle with period 20 $=\phi(25)$. Part 1 is incorrect with $a_{n+100}$, since $2^{100} \equiv 1 \pmod{25}$. $\endgroup$
    – Calvin Lin
    Oct 4 '19 at 3:29
  • $\begingroup$ @CalvinLin: you are correct and my second paragraph supports that. Fixed. $\endgroup$ Oct 4 '19 at 3:32

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