Boeing717
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| Joined: 08 Jun 2008 |
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| 28 Feb 2012 07:19 PM |
| Are there whole numbers which can not be divided by 9, but if multiplied by two, are multiples of 9? |
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| 28 Feb 2012 07:20 PM |
can you re-word that?
Friendship is a potato |
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Jacob147
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| Joined: 24 Feb 2008 |
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| 28 Feb 2012 07:20 PM |
hmmmmmmmmmmmmmmmmmm im going to explode nao |
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| 28 Feb 2012 07:20 PM |
| wait never mind, i'm thinking of something else |
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chaser67
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Boeing717
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| Joined: 08 Jun 2008 |
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| 28 Feb 2012 07:21 PM |
| Okay, imagine we have an integer x. Now we multiply by 2, getting integer 2x. If 2x is a multiple of 9, is there any value from which x would not be a multiple of 9? |
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| 28 Feb 2012 07:22 PM |
| Technically, only 4.5 would fit this rule. Well, as far as I can tell. |
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Jacob147
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Boeing717
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| 28 Feb 2012 07:23 PM |
even
I said a whole number. |
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| 28 Feb 2012 07:23 PM |
| Wait a sec... That wasn't a whole number. *facepalm* |
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| 28 Feb 2012 07:24 PM |
| I can't name any other numbers relating to this theory at all... I'm too lazy to go to my living room and get my calculator, so I'm out of here... |
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| 28 Feb 2012 07:25 PM |
interesting thought though but I am pretty sure the answer is no
you can find a multiple of nine by if it's digits all equal another multiple of nine
so if you manage to get something like that without multiplying anything by nine, I guess so
but otherwise, as I said before, no
~f٥гﻉ√ﻉг™~ |
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| 28 Feb 2012 07:27 PM |
| Any number can be divided by 9, evenly or not, so no. |
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Boeing717
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| Joined: 08 Jun 2008 |
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| 28 Feb 2012 07:28 PM |
It can actually be extended further.
If an even number has an odd divisor, then half that number can be divided by the same integers. (i.e for 18, you can divide by the odd integers 1, 3, and 9, divided by 2, we get 9, which can be divided into the same 3)
Anyone who wants to try a proof or a refutation to this can. I have a feeling the converse is true, where no even number can be divided by an odd number that half of it could (in the same case as the original, if we have 9 and multiply by 2, we won't be able to divide evenly into 7 or 5 or anything like that). |
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