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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?

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    $\begingroup$ Yeah, $x=125$ definitely seems wrong. $\endgroup$
    – Thomas Andrews
    Nov 15, 2018 at 22:47
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    $\begingroup$ 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 $x=3$. $\endgroup$
    – Sangchul Lee
    Nov 15, 2018 at 22:57

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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.

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