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 #18
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In your original method, you suggested that the number matrix be filled out with leading zeros so that all the numbers have the same width. In a set of integers between 0 and 1 * 10^12, this would create a square matrix with zeros clustered at the uper left, like this:

 

    000000000000

    000000000003

    000000000014

    000000000481

    000000004067

    000000011202

    000000765025

    000006570349

    000052570231

    000569504785

    007221909937

    010323716328

    838219253314

 

This change makes it easier to understand the problem, but it doesn't change the result, which is true for a set of sequential 1 * 10^12 integers that aren't padded in this way:

 

 

                                     0

                                     6

                                   46

                                 426

                               1435

                             55123

                           230105

                         2873680

                       63725050

                     640899375

                   0060833473

                 53442421828

               372859331893

 

The point? The differing length of the integers between 0 and 1 * 10^12 needs to be taken into account. So the number of threes in the set of integers between 0 and 1 * 10^12 is 1 * 10^12 (height) times 12 (width) divided by 10 (isolate any single digit) divided by two (to account for the missing digits in the upper left of the martix as shown above).

Apr 11, 2017
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 #1
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Apr 10, 2017
 #2
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