Lets say B is a natural number with n digits. Some of the digits of B are zero. C is a natural number obtained by multiplying B with A. $$C=A\times B$$ Can you always find the smallest natural number A such that the number C doesn't have any digits which are zero? Is there any research done in this topic? Is this just trial and error?
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1Of course you'll need that the last digit of $B$ is not zero. – aschepler Jun 28 '23 at 21:59
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1Well, if $10,|,B$ then no such $A$ exists. If $\gcd(10,B)=1$ then $10^{\varphi(B)}-1$ is a string of $9's$ and it is divisible by $B$. That probably won't be minimal, but at least it gives an upper bound to your search. That leaves the cases in which $2$ or $5$ (but not both) divide $B$. – lulu Jun 28 '23 at 21:59
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How do you know that $$10^(phi(B))-1$$ is a multiple of B? I think that information will be helpful to me. Thank you for your time. – Robert Puscasu Jun 28 '23 at 22:08
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1https://math.stackexchange.com/questions/2067752 shows existence. – aschepler Jun 28 '23 at 22:09
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@aschepler Oh, thank you. I was trying to find that. – lulu Jun 28 '23 at 22:09
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1@lulu It was in the automatically generated "Related" questions list. ;) – aschepler Jun 28 '23 at 22:10
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@RobertPuscasu That's just Euler's Theorem – lulu Jun 28 '23 at 22:10
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1@aschepler Ah. It's been pointed out to me before that I ought to sometimes look at that list. Seems like good advice. – lulu Jun 28 '23 at 22:11