Re^2: Can It Be Written In Base X With Only 1s And 0s

[Laurent_R]

82000 is a sum of powers of all the inputs

That's sort of by definition: if the number is expressed with only 1 and 0 in a given base, then it will be sum of simple powers of that base.

[BrowserUk]

Hm. A phraseology problem I think.

Below are all the numbers less than 3^11, that when encoded base-3 use only 0 1:

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All of them can be defined as the sums of multiples of powers of 3. But only the first number in each block is a power of 3.

And 82000 is a sum of a selection of those first numbers in each block. And that is so for 4 & 5 also.

And, if it holds true for the higher numbers in the sequence (and they are going to be very large) then not having to consider all the other numbers in each of those blocks is a significant saving.

So worth pointing out don't you think? Even if I need to clarify the meaning or use better phraseology.

How about a sum of single powers? Or discrete powers?

[Laurent_R]

Yes, it is probably a phraseology problem or a misunderstanding, but I still don't agree with this sentence:

All of them can be defined as the sums of multiples of powers of 3. But only the first number in each block is a power of 3.

To me, these numbers are all sums of single powers of 3. For example, taking the beginning of your list:

1          -- 3**0
3 4        -- 3**1, 3**0 + 3**1
9 10 12 13 -- 3**2, 3**0 + 3**2, 3**1 + 3**2, 3**0 + 3**1 + 3**2
etc.

So they are all sums of pure (or single) powers of 3, not sums of multiples of powers of 3 (which would imply numbers expressed with other digits than 0 and 1 in base 3). And so is 82000.

And I agree with you that you don't have to consider these other numbers, only those that are pure powers of 3 are of interest for the search; so, as you said, only the first one of each block if you want to figure out whether 82000 or any other number qualifies the test.

[BrowserUk]

82000 can be broken down to 3^0 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^9 + 3^10. Ie. The sum of 8 discrete or single powers of 3.

82001 would require 2*3^0, thus a multiple of one of the powers of 3.

I don't much care how it is worded; so long as you understand my meaning; which you evidently do.

[ikegami]

All of them can be defined as the sums of multiples of powers of 3.

All integers are sums of multiples of powers of 3. The ones of interest are sums of powers of 3. You had it right the first time

[BrowserUk]

Jeez!

These are all powers of 3 less than 3^11:

1 3 9 27 81 243 729 2187 6561 19683 59049

These are all sums of multiple powers of three less than 3^11 that also consist of 0s & 1s when encoded in base-3:

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It is a quicker to permute the combinations of the 11 powers of the to look for solutions that are sum of distinct powers of 3,

than it is to permute the combinations 1847 compliant numbers that are (a subset of the) sums of multiple powers of 3.

And the numbers & differences get much larger for 4, 5, 6 ...

So, please refrain from telling me what I got right and wrong, when you appear to not understand the subject.