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Topic: The non-negative integers (Read 5065 times) |
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wdcefv
Newbie
Posts: 1
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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
Senior Riddler
Posts: 460
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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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Posts: 735
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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?
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| « Last Edit: Jul 28th, 2014, 6:25pm by dudiobugtron » |
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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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