Divisors

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User2020

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I recently read that if a number x is not divisable by 2 and 5 then there is such a number whose digits are even, odd, even, odd... or odd, even... and this number is divisable by this number x. How do I proof that?
 
What do you know about x?
Well, if a number is not divisible by 2, than it is not even, and also cannot be divided by any other even number. Therefore, x is not even. If it cannot be divided by 5, then...well, it can't be divided by 5! Three is still open to us, however, but 6 is not. Do you see a pattern?

Try to work with that hypothesis.
 
User2020 said:
I recently read that if a number x is not divisable by 2 and 5 then there is such a number whose digits are even, odd, even, odd... or odd, even... and this number is divisable by this number x. How do I proof that?
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The first thing to do is to clarify the problem. Does "not divisible by 2 and 5" mean "not divisible by both 2 and 5" (that is, not divisible by 10), or "not divisible by 2 and not divisible by 5" (that is, not divisible by either 2 or 5"), or something else?

I'll take it as meaning this:

If x is not a multiple of 10, then there is a multiple of x whose digits are alternately odd and even (starting with either an odd or an even digit).​

It would be best if you could show us the original wording of the claim.

Also, we'll to know your level of knowledge of the subject, starting with what ideas you have for proving the claim (assuming, of course, that it is even true).
 
I mean it's not a multiple of 2 or 5. For example number like 1836 is a multiple of 3 and 9 and it's digits are odd, even, odd, even. I think that maybe I should find a number like 1000 or 2000 and to who I can add another number so that those two numbers are not divisable by x, but their sum is.
 
User2020 said:
I mean it's not a multiple of 2 or 5. For example number like 1836 is a multiple of 3 and 9 and it's digits are odd, even, odd, even. I think that maybe I should find a number like 1000 or 2000 and to who I can add another number so that those two numbers are not divisable by x, but their sum is.
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That doesn't really clarify things.

The claim was,

If a number x is not divisible by 2 and 5 then there is such a number whose digits are even, odd, even, odd... or odd, even... and this number is divisible by this number x.​

What is x in your example? [MATH]1836 = 2^2\cdot 3^3\cdot 17[/MATH], so it could be an example of a multiple of 2, or of 3, or of 17, that fits the pattern. But this is a multiple of 2, so how does it fit the problem, if the claim is about a number that is not divisible by either 2 or 5?

I'm now even less sure what the claim really is. I really need to see the original wording (even if it was in another language!).
 
Darn..this is confusing.
Otis said:
Hello firemath. We can say that x does not end in 5 or 0.
Anyone who remotely knows their times tables does not want to have their intelligence insulted by me telling them that the number does not end in 5 or 0....Or so I think.:unsure:


What hypothesis?
Do all the numbers that I took out of the picture count as a hypothesis?



\(\;\)
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Jomo said:
You can say it does not end in 5 (since we already know the number is not even)
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Were you addressing @Otis?
 
firemath said:
Anyone who remotely knows their times tables does not want to have their intelligence insulted by me telling them that the number does not end in 5 or 0 …
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So you told them instead, "If it cannot be divided by 5, then...well, it can't be divided by 5!" (By the way, specifically knowing the last digit is not 5 or 0 may play a role in somewhere in a proof.)

I'm more interested in your hypothesis. Can you state it?

?
 
I read the op to mean:

If a number x is not divisible by 2 or 5, then there is a multiple of x called y whose digits are [even,odd,even,odd,…] or [odd,even,odd,even,…]

Example:

x = 1449 = 32 · 7 · 23

y = 4347 (3rd multiple), 10143 (7th multiple), etc

?
 
What intrigues me is that I've tried finding such a multiple for every number x up to 100, and the only numbers for which I can't find one are multiples of 20. Other even numbers or multiples of 5 work fine! So if the problem is really what it appears to be, the condition is far from necessary; it may be included only in order to make the proof easier. But I can't see a way to prove it under any condition. I don't know what technique can apply to such a question.

That's part of the reason I'm unsure that we're reading the question correctly. (The grammar isn't exactly clear in the first place.)
 
firemath said:
I would die before I learned how to read their notation.
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Some of the statements of these open problems are easy enough for a high school algebra student to understand. You have graduated high school, right?:)
 
lookagain said:
I don't hate "unsolved problems." This was an unfinished problem.
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To me, the trouble is that it is not even clear what it means! In that sense, I'd call it "unresolved". It was stated as if the (apparently unnecessary) condition would make it easier to prove, but we're left hanging.
 
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