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Topic: Powers of 2 (Read 2585 times) |
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NickH
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Powers of 2
« on: Aug 28th, 2004, 11:45am » |
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Does there exist an integral power of 2 such that it is possible to rearrange the digits giving another power of 2?
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Barukh
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Re: Powers of 2
« Reply #1 on: Aug 29th, 2004, 1:48am » |
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[smiley=blacksquare.gif]
No: If A = 2n > A’ = 2k, then 9|(A-A’), but A-A’ = 2k(2n-k-1), and n-k [le] 3.
[smiley=blacksquare.gif]
That's not correct...
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| « Last Edit: Aug 29th, 2004, 3:33am by Barukh » |
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ThudnBlunder
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Re: Powers of 2
« Reply #2 on: Aug 29th, 2004, 4:10am » |
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Let n1 = 2n+r
Let n2 = 2n
And 2n+r - 2n = 2n(2r - 1) = 9k if we are using decimal.
So 9 must divide 2r - 1
But, if n1 and n2 have the same number of digits, then r = 1, 2, or 3
Hence no such decimal number exists.
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| « Last Edit: Aug 29th, 2004, 1:29pm by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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NickH
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Re: Powers of 2
« Reply #3 on: Aug 29th, 2004, 5:36am » |
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That's the method I used, T&B.
What if leading zeroes are allowed in the (decimal) numbers?
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Barukh
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Re: Powers of 2
« Reply #4 on: Aug 29th, 2004, 5:38am » |
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on Aug 29th, 2004, 5:36am, NickH wrote:That's the method I used, T&B.
What if leading zeroes are allowed in the (decimal) numbers? |
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Hmm, I used essentially the same method, but then realized that it may not work because of the leading zeros ;D
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Disoriented
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Re: Powers of 2
« Reply #5 on: Aug 31st, 2004, 5:03pm » |
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on Aug 29th, 2004, 4:10am, THUDandBLUNDER wrote:Let n1 = 2n+r
Let n2 = 2n
And 2n+r - 2n = 2n(2r - 1) = 9k if we are using decimal.
So 9 must divide 2r - 1
But, if n1 and n2 have the same number of digits, then r = 1, 2, or 3
Hence no such decimal number exists.
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I lost you on line 3. Could you explain in a way that makes sense to non-math majors?
Thanks 
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asterix
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From one non-math major to another, I never would have figured out the solution, but I think I can decode the answer. First, if you take any number and rearrange the digits the difference between the two will be divisible by 9. (for example, with the 2 digits a and b, ab-ba=(10a+b)-(10b+a)=9a-9b, which must be divisible by 9).
So 2n+r=2n*2r so the difference between our two numbers will be 2n*(2r-1).
It must be divisible by 9, but since 2n is never divisible by 9, then 2r-1 has to be.
Did I get that right?
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ThudnBlunder
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Re: Powers of 2
« Reply #7 on: Aug 31st, 2004, 10:42pm » |
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Quote:
Yep.
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