Puzzle time - integers
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wrote on 20 Jul 2020, 09:54 last edited by
I'm happy to help
There are many such definitions where you want something different from "smallest" (e.g. coinductive definitions).
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wrote on 20 Jul 2020, 09:58 last edited by Klaus
1 is not in S.
(note that you didn't ask for an exhaustive list of those not in S
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wrote on 20 Jul 2020, 10:02 last edited by
...and if you want the complete set:
It's the set of positive integers minus S.
Which is a perfectly valid mathematical definition of the integers not in S.
So presumably you want us to specify that set in a particular way?
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wrote on 20 Jul 2020, 10:06 last edited by
Ha
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wrote on 20 Jul 2020, 11:27 last edited by Klaus
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Obviously, when a number n is in S, then n+5 must also be in S.
So once we have all digits from 0 to 4 (or 5 to 9) as last digits of numbers, all numbers above it must be in S.
So the question is whether we ever get all last digits.
I think we can get to all last-digits except 0 and 5, since any number that ends with 0 or 5 squared also ends with 0 or 5.
So, my theory about the positive integers not in S is:
There's some noise in the beginning, and after a while it's only the numbers that end with 0 or 5.
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wrote on 20 Jul 2020, 11:29 last edited by
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On the right track but not quite there
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wrote on 20 Jul 2020, 11:32 last edited by
So you are saying my last statement is wrong, or are you saying it's not precise enough?
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wrote on 20 Jul 2020, 11:37 last edited by
Depends on how one defines ‘noise‘. But what I really mean is “from what I infer from your words you’re still missing an insight here”
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wrote on 20 Jul 2020, 11:59 last edited by
OK, here's a precise version of the statement:
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There is a number N, such that for all n >N, n is not in S if and only if the last digit of n is 0 or 5.
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Is that correct?
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wrote on 20 Jul 2020, 12:02 last edited by jon-nyc
Yes but tell me N. You’re missing something or you would know what N is.
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wrote on 20 Jul 2020, 12:10 last edited by Klaus
N is smaller than or equal to 2915. Now don't tell me you want me to worry about selecting a particular number between 1 and 2915!!!
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wrote on 20 Jul 2020, 13:10 last edited by
Yes I do.
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wrote on 20 Jul 2020, 18:14 last edited by
What Klaus missed:
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The only ‘noise’ (besides all multiples of 5) is the number 1.
- 2 is granted which gets you all numbers ending in 2 or 7.
- 7^2 is 49 which gets you all the numbers ending in 9 and 4 above that
- after 49 is 54. 54^2 is 3136 which gets you all the numbers ending in 6 or 1 above it.
BUT - once you have the *6s, you’ll get to 6^8 which gets you back to 6 and 11, etc.
- that gets you to 16 which gets you back to 4 and 9
- that 9 gets you back to 3 and 8
So we have 2,3,4,6,7,8,9 covered plus any number that is a multiple of 5 above them.
So only 1 is missing, along with all multiples of 5
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wrote on 20 Jul 2020, 20:52 last edited by
Nice!
54^2 is 2916 and not 3136, though - that was the source of the 2915 bound I was giving above. So my bound was pointing in the right direction
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wrote on 20 Jul 2020, 20:58 last edited by
My math buddy at CS pointed out that Fermat’s Little Theorem could help here too rather than finding actual paths back to the lower numbers.