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BNC
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divide by 17
« on: Apr 5th, 2003, 11:17pm » |
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What is the smallest natural number that:
1. The sum of its digits is divisable by 17, and
2. The sum of the following number's digits also divisable by 17
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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LZJ
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Re: divide by 17
« Reply #1 on: Apr 6th, 2003, 1:32am » |
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8899?
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BNC
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Re: divide by 17
« Reply #2 on: Apr 6th, 2003, 7:16am » |
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Yes.
How did you get it?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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LZJ
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Re: divide by 17
« Reply #3 on: Apr 6th, 2003, 7:54am » |
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Well, logic dictates that the 2 numbers have to end off with 99 and 00, but not 999 and 000, so that when modulus 17 is taken, they will have the same remainder. The smallest possible natural number of mod 17 = 0 with this property is 8899.
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Noke Lieu
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Re: divide by 17
« Reply #4 on: Mar 7th, 2004, 7:54pm » |
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Just to be picky,
1 is divisible by 17. You get one seventeenth.
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rmsgrey
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Re: divide by 17
« Reply #5 on: Mar 8th, 2004, 5:20am » |
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on Mar 7th, 2004, 7:54pm, Noke Lieu wrote:Just to be picky,
1 is divisible by 17. You get one seventeenth.
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It depends on what set of numbers you're retricting yourself to - in [bbn] or [bbz] only some numbers are divisible by 17. In any field including 17, everything is divisible by 17.
Since the question mentions natural numbers, it seems natural to restrict ourselves to functions [bbn][to][bbn]
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Sameer
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Re: divide by 17
« Reply #6 on: Mar 8th, 2004, 2:38pm » |
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on Apr 6th, 2003, 7:54am, LZJ wrote:| Well, logic dictates that the 2 numbers have to end off with 99 and 00, but not 999 and 000, so that when modulus 17 is taken, they will have the same remainder. The smallest possible natural number of mod 17 = 0 with this property is 8899. |
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Can you prove this? I am unable to get any answers by calculations!!!
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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rmsgrey
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Re: divide by 17
« Reply #7 on: Mar 9th, 2004, 4:17am » |
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Let f(x)=Sum of x's digits.
Then:
f(x+1)=f(x)+1-9n where x=10n-1 (mod 10n) and x[ne]10n+1-1 (mod 10n+1)
In other words, if x ends with a string of n 9's, then the sum of its digits gets reduced by 9n-1 when x increases by 1.
Let g(x)=f(x) (mod 17)
We want min{x[epsilon][bbn]: g(x)=g(x+1)=0}
For g(x)=g(x+1), we have 9n-1=0 (mod 17) giving n=2 (mod 17) so x must end in 99 but not 999 unless it goes on to end in 9,999,999,999,999,999,999...
g(99)=1, so we want the minimum number, y, to prefix to 99 that satisfies:
(i) g(y)=16
(ii) y[ne]9 (mod 10)
There is obviously no one digit y, and in two digits, 79 ends in a 9, leaving 88 as the least possible prefix for the 99 giving x=8899, x+1=8900.
Hopefully that fills in the gaps in LZJ's working sufficiently well for anyone interested to fill in any remaining gaps themselves.
[e]Purely cosmetic[/e]
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| « Last Edit: Mar 9th, 2004, 4:19am by rmsgrey » |
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Sameer
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Re: divide by 17
« Reply #8 on: Mar 9th, 2004, 6:50am » |
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Absolutely amazing rmsgrey.
However, would we be able to prove the number itself. E.g. your proof assumes we already have string of n 9s at end. Is there a proof to actually get the number itself or lead to a conclusion that there has to be n 9s at the end?
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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rmsgrey
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Re: divide by 17
« Reply #9 on: Mar 10th, 2004, 4:08am » |
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My proof assumes there is a string of n[ge]0 9s at the end. If you can find a number in [bbn] that doesn't end in a string of at least 0 9s, I'll willingly try and correct my proof.
[e]Sorry, re-read the written version of my proof, and it turns out my "purely cosmetic" revision dropped a case[/e]
[e2]Re-re-reading throws up the question of what (mod 1) means in context and probably x=100-1 (mod 100 holds for all x[epsilon][bbn]. Certainly the written paraphrase holds.[/e2]
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| « Last Edit: Mar 10th, 2004, 4:15am by rmsgrey » |
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