# If $2^n=(... \underbrace{aa...aaa}_{x\text{ times}})_{10}$ what is maximum of $x$?

If $$2^n=(... \underbrace{aa...aaa}_{x\text{ times}})_{10}$$ for positive integer $$n$$ then we have

$$2^n=(... \underbrace{aa...aaa}_{x\text{ times}})_{10}=(b\underbrace{aa...aaa}_{x\text{ times}})_{10}=10^x b+a(10^{x-1}+...+1)$$ Where $$a$$ is a digit between $$0$$ and $$9$$ and $$b$$ is a natural number.

Simplifying more gives $$9 \times2^n=9b \times 10^x + a (10^x-1)$$

Now if $$n>3$$ and $$x>3$$ then taking mod 2 gives $$v_2(a)>3$$ which is impossible. So one $$x$$ or $$n$$ is not greater than $$3$$. But the book says answer is $$x=125$$. Am I interpret this wrong?

• Yeah, $x=125$ definitely seems wrong. Nov 15, 2018 at 22:47
• I see that your solution is right. For a different approach, one can prove that $2^{n+500} \equiv 2^n \pmod{10^4}$ for all $n \geq 4$ (by using $\varphi(5^4)=500$, for instance), and up to $n \leq 503$, $x$ only takes the values $\{1, 2, 3\}$, with $2^{39} = 549755813888$ the smallest case with $x=3$. Nov 15, 2018 at 22:57

If $$n\ge 4$$, then $$2^n$$ must be a multiple of $$16$$, therefore so are its last four digits. There are no factors of $$5$$ so the last four digits cannot be $$0000$$. Now, which other possibilities for four identical last digits will give a multiple of $$16$$? Try them out and see.
The comments identify $$2^{39}=...888$$. If you work out the part above you are left with what must be the answer.