Integer Divisible by 2^2009 with No Zero Digits

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Integer Divisible by 2^2009 with No Zero Digits

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zeta1

#1 h Apr 9, 2025, 5:57 PM

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Show that there exists a positive integer that has no zero digits and is divisible by 2^2009.

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#2 h Apr 9, 2025, 6:51 PM • 1 Y

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First, we prove by induction that there is a positive integer divisible by $2^{2009}$ where the last $n$ digits are not 0, for all positive integers $n$ . Tag this claim (*).

Base case: (*) holds for $n=1$
Proof: $2^{2009}$ is not divisible by $10 = 2 \cdot 5$ , therefore its last digit is not 0.
Induction case: (*) holds for $n=k$ implies (*) holds for $n=k+1$
Proof: Let $a_k$ be a positive integer divisible by $2^{2009}$ such that its last $k$ digits are not 0. If the $k+1$ th last digit of $a_k$ is not 0, the goal holds. Otherwise, the $k+1$ th last digit of $a_k$ is 0. Consider the number $a_k(10^k + 1)$ . This number has the same last $k$ digits as $a_k$ , and the $k+1$ th last digit of $a_k(10^k+1)$ is the same as the last digit of $a_k$ , and is therefore not 0. As such, the last $k+1$ digits of $a_k(10^k + 1)$ are nonzero.

With that done, let $a_{2009}$ be a positive integer such that its last $2009$ digits are nonzero. Notice that $10^{2009}$ is divisible by $2^{2009}$ , so $a_{2009} + b \cdot 10^{2009}$ is divisible by $2^{2009}$ for any nonnegative integer $b$ . As such, any sufficiently large number with the same last 2009 digits as $a_{2009}$ is divisible by $2^{2009}$ . There exists a sufficiently large number with no zero digits and that has the same last 2009 digits as $a_{2009}$ , as desired.

This post has been edited 2 times. Last edited by maromex, Apr 9, 2025, 6:53 PM
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