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Topic: The nonnegative integers (Read 3013 times) 

wdcefv
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Posts: 1


The nonnegative integers
« on: Jun 13^{th}, 2014, 4:58am » 
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The nonnegative 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
Senior Riddler
Posts: 460


Re: The nonnegative integers
« Reply #1 on: Jul 28^{th}, 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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Posts: 689


Re: The nonnegative integers
« Reply #2 on: Jul 28^{th}, 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 ndigit 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 28^{th}, 2014, 6:25pm by dudiobugtron » 
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gotit
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Re: The nonnegative integers
« Reply #3 on: Jul 29^{th}, 2014, 2:20am » 
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Here is my quick calculation For a Ndigit nonnegative integer, where N > 1 n(A) = 4 * 5^{N1} Reason: A can contain numbers that have only 0,3,6,8 or 9 as its digits n(B) = 3^{N} Reason: B can contain numbers that have only 1,4,7 as its digits n(C) = 10^{N}  10^{N1}  (n(A) + n(B)) Reason: Subtract n(A) + n(B) from the number of possible Ndigit 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 nonnegative integers
« Reply #4 on: Jul 29^{th}, 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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