# Find last n for which 2^n has a 0.

Find last number $n$ for which $2^n$ has a zero. For example $2^{10}=1024$ has a zero for which last number zero will be there. (It is possible that there doesn't exist such limits to $n$ but what is the proof?)

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Is this homework? What have you tried? –  Mike Miller Apr 20 '14 at 6:58
How are you defining last $n$? Additionally, how do you handle fractional cases as $2^{-12}=0.00024414062$ that has a 0 and is a rather small number compared to others. –  JB King Apr 20 '14 at 8:12

By Euler's theorem, $2^{20k}\equiv1\pmod{25}\implies 2^{20k+2}\equiv04\pmod{100}$, so..

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(L=Little, contrary to L=Last.) –  Asaf Karagila Apr 20 '14 at 7:40
This is such a far better approach than my own it's not even funny. –  Mike Miller Apr 20 '14 at 7:43
That's brilliant!What was your motivation behind this solution?Did you first think about showing that there are infinitely many powers of two with zero as the second-to-last digit?Also,can we not comment using Euler's totient theorem that $2^{40k}\equiv 1\pmod {100}$ and the result follows? –  rah4927 Apr 20 '14 at 7:59
Hehe, you're welcome! –  Asaf Karagila Apr 20 '14 at 8:05
@rah4927 Just as a note: the result you'd hoped for there is very silly, since the left-hand side is a power of two, and the right-hand side says that our number is odd! –  Mike Miller Apr 20 '14 at 8:12

Hint: Prove that infinitely many $2^n$ will start with $10$.

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How to prove this? I am confused. –  Satvik Mashkaria Apr 20 '14 at 7:02
Since $\log_{10}2$ is irrational, the set $\{\{n\log_{10}2\}:n\in\mathbb Z^+\}$ is dense in $[0,1]$. (You can prove this with the Pigeon Hole Principle for any irrational number.) Knowing this, try to show that $2^n$ can start with $10$ when written in decimal. –  barto Apr 20 '14 at 7:22
@barto: Using $\{x\}$ to denote the fractional part while using $\{\ldots\}$ to denote a set is an effective example of when operator overloading is bad. Perhaps writing $x-\lfloor x\rfloor$ is better for readability. –  Asaf Karagila Apr 20 '14 at 7:42
@barto: I agree, and I understand your plight. But that's life... :\ –  Asaf Karagila Apr 20 '14 at 8:10
@barto I've occasionally seen $\mod 1$ used as shorthand for the fractional part. –  Mike Miller Apr 20 '14 at 8:12