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).

  • 6
    $\begingroup$ 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

$2^{30}$ 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.

| cite | improve this answer | |
  • 5
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.