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   The non-negative integers
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   Author  Topic: The non-negative integers  (Read 3013 times)
wdcefv
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The non-negative integers  
« on: Jun 13th, 2014, 4:58am »
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The non-negative integers are divided into three groups as follows:
A= {0,3,6,8,9,...}, B= {1,4,7,11,14,...}, C= {2,5,10,13,...}
Explain.
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JohanC
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Re: The non-negative integers  
« Reply #1 on: Jul 28th, 2014, 3:10pm »
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Very interesting.
Maybe you first tried to straighten some numbers, and then got to round some other numbers. But you were left with a whole bunch that is neither the one nor the other, neither fish nor fowl.
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dudiobugtron
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Re: The non-negative integers  
« Reply #2 on: Jul 28th, 2014, 6:24pm »
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Awesome riddle wdcefv, and great deduction JohanC.  I had thought at length about this puzzle and gotten nowhere; I'd in fact given it up for unsolvable.
 
If JohanC's solution is the intended one, then it raises some interesting questions about the relative sizes of the sets.  Obviously Set C will grow at an increasingly faster rate as you keep adding numbers.  It also seems that, for some random n-digit number, the chance that it is in Set A or B decreases as n increases.  My question is - what are the relative sizes (measures?) of these sets?  I understand all three sets are infinitive, but is there a way to compare them that shows whether Set A and B are negligible in size compared to C?
« Last Edit: Jul 28th, 2014, 6:25pm by dudiobugtron » IP Logged
gotit
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Re: The non-negative integers  
« Reply #3 on: Jul 29th, 2014, 2:20am »
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Here is my quick calculation
 
For a N-digit non-negative integer, where N > 1
 
n(A) = 4 * 5N-1
Reason: A can contain numbers that have only 0,3,6,8 or 9 as its digits
 
n(B) = 3N
Reason: B can contain numbers that have only 1,4,7 as its digits
 
n(C) = 10N - 10N-1 - (n(A) + n(B))
Reason: Subtract n(A) + n(B) from the number of possible N-digit integers.
 
So as N becomes bigger, n(A) and n(B) will become negligible compared to n(C).
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rmsgrey
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Re: The non-negative integers  
« Reply #4 on: Jul 29th, 2014, 7:28am »
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It's a known result that practically every number has a 5 among its digits - more formally: for any probability p>0, there exists an N, such that for all n>N, when you select a single number uniformly at random from all positive integers up to n, the probability that there is no 5 among its digits is less than p.
 
That |A+B| / |C| -> 0 follows trivially from that result.
 
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