"All base 3 numbers above a certain size... have a 0 in them."

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jewish_scientist
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"All base 3 numbers above a certain size... have a 0 in them."

Postby jewish_scientist » Tue Jun 04, 2019 6:59 am UTC

In this Numberphile video, the professor says, "There is a conjecture that all base 3 numbers above a certain size, I forget what it is, have a 0 in them." I do not understand this conjecture. Can't I always take any base 3 number that has a 0 and substitute the 1s for the 0s and have another valid base 3 number?
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ConMan
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Re: "All base 3 numbers above a certain size... have a 0 in them."

Postby ConMan » Tue Jun 04, 2019 7:20 am UTC

I believe he's talking specifically about the property of multiplicative persistence - so my suspicion is that the conjecture he's trying to describe is that all base 3 above a certain size, when you multiply the digits together will result in a new number that has a 0 in them, and hence all base 3 numbers above a certain size have a multiplicative persistence of 2.
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cyanyoshi
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Re: "All base 3 numbers above a certain size... have a 0 in them."

Postby cyanyoshi » Tue Jun 04, 2019 7:25 am UTC

Well, obviously you can write arbitrarily large numbers in base 3 without any zeros. He had to have misspoken.

But here's what I think he was getting at. In base 3, the only digits are 0, 1, and 2.* If a number has a 0 anywhere, then the product of its digits is 0. Otherwise, this product is a power of 2. So how do you write the powers of 2 in base 3? Well,

1 -> 1
2 -> 2
4 -> 11
8 -> 22
16 -> 121
32 -> 1012
64 -> 2101
128 -> 11202
256 -> 100111

I think what he meant to say is that all powers of 2 above a certain size must have a 0 in them somewhere when written in base 3, so you'd only have finitely many cases to check.


*Unless you consider other digits, of course.


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