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    The number 1920212223...939495 consists of 154 digits, and was arrived at by placing the numbers from 19 to 95 one after the other. If I now remove 95 of the digits in such a way that the resulting number is the greatest possible, what are the first digits of the 59-digit number that remains?

    Question #23745. Asked by Math Matt. (Oct 31 02 2:18 PM)


    Bryce

    		For the resulting number to be as great as possible it is necessary for it to have the highest number of nines to start with. If all the digits other than those equal to nine are removed from the original number, from the beginning up to 5 in 59, 77 digitswill have been removed, and 5 nines will remain. If we now remove all the digits from the 60 to 67 as well as the 6 in 68 (that is, a total of 17 digits), 94 digits in total will have been removed and what is left is a number that begins with the digits 9,9,9,9,9, and 8. The 6 in 69 is now removed, bringing us to the last of the 95 to be removed.
    What we are left with is therefore a number whose first 19 digits are 9999989707172737475.

    Oct 31 02, 5:28 PM

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