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Thread: Perfect Square Number

  1. #1
    wps
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    Perfect Square Number

    $N$ is written in base $10$, and the digits in $N$ are given by exactly $600$ digits "$1$", with all the other digits being "$0$". Can $N$ be a perfect square?
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    Re: Perfect Square Number

    $N=10^{599} = 10^{598}\cdot 10$

    $\sqrt{N} = \sqrt{10^{598}\cdot 10} = 10^{299}\sqrt{10}$

    $N$ is not a perfect square
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    Re: Perfect Square Number

    I interpret the question a little differently. N is a positive integer written in base 10 where the only decimal digits are either 0 or 1. (So N may have any number of digits greater than or equal to 600, say 2011 total digits.) Remember ("casting out nines"), N is congruent modulo 9 to the sum of digits of N. This sum is then 600. Since 600 is divisible by 3, N is divisible by 3, but since 600 is not divisible by 9, $3^2$ does not divide N. Hence N can not be a square.
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    Re: Perfect Square Number

    your interpretation is certainly more interesting
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