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Prove that $2^{30}$ has at least two repeated digits.

I assume that the question is asking me to prove that $2^{30}$ has at least one digit that appears twice. Correct me if I'm wrong. (I later checked $2^{30}$ has three digits each of which appears twice, I initially thought that if I could prove the $2^{30}$ has 11 digits, then I can prove the given, but calculated the number of digits only to find out that it has 10 digits).

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    $\begingroup$ Hint: What is $2^{30} \pmod 9$? $\endgroup$ – user670344 May 15 '19 at 18:14
  • $\begingroup$ $2^{30}=1073741824$, so ... $\endgroup$ – Sil May 19 '19 at 17:10
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$2^{30}$ has $10$ decimal digits, as $10^9 < 2^{30} < 10^{10}$. If none were repeated, each of the $10$ digits $0$ to $9$ would appear once. But if that were the case, the sum of digits would be $45$, which would make the number divisible by $9$, and $2^{30}$ is not.

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    $\begingroup$ This answer seems to depend on convincing yourself that $2^{30}$ has 10 digits without actually multiplying it out or resorting to logarithms (since doing that would violate the spirit of the question). The programmer in me knows that $2^{10}$ is $1024$. So $2^{30} = (2^{10})^3$ is a little more than $1000^3$, which is easy to see has 10 digits. $\endgroup$ – csd May 15 '19 at 18:57

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