# "Convergence" of the sequence $a_k=2^{10^{\ k}}$

I've been observing final digits of each number in the sequence $$a_k=2^{10^{\ k}}$$

You get:

$\ a_0=2 \\ a_1=1024 \\ a_2= ...205376 \\a_3= ...069376\\a_4=...709376\\a_5=...9883109376\\a_6=...2747109376\\a_7=...1387109376$

And so on. Obviously, the numbers are getting exponentially larger, but the final digits of each number appear to be "converging".

Purely out of curiosity, I'm wondering if anyone can explain this effect.

• $2^{100}=6$ mod $10$, and $6\times 6=6$ mod $10$, so $2^{k}=6$ mod $10$ whenever $k$ is a multiple of $100$. Oct 21 '13 at 19:33

By Euler's generalisation of Fermat's theorem, we have

$$a^{\varphi(n)} \equiv 1 \pmod{n}$$

whenever $a$ and $n$ are coprime, where $\varphi$ is Euler's totient function. For the base $a = 2$ and the modulus $n = 5^m$, we have $\varphi(5^m) = (5-1)\cdot 5^{m-1}$, so, for $k \geqslant 2$ we have

$$2^{10^k} = \left(2^{4\cdot 5^k}\right)^{2^{k-2}} \equiv 1 \pmod{5^{k+1}}.$$

And we have $10^k \geqslant k+1$ for all $k \geqslant 0$, so

$$2^{10^k} \equiv 0 \pmod{2^{k+1}}.$$

Thus the last $k+1$ digits of $2^{10^k}$ are the unique (modulo $10^{k+1}$) solution to

$$x_k \equiv 1 \pmod{5^{k+1}}\land x_k \equiv 0 \pmod{2^{k+1}}.$$

Obviously, we also have

$$x_k \equiv 1 \pmod{5^k}\land x_k \equiv 0 \pmod{2^k},$$

so $x_k \equiv x_{k-1} \pmod{10^k}$, which means that (for $k \geqslant 2$) the last $k+1$ digits don't change anymore.