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  1. #1
    Newbie kira's Avatar
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    Post Show that

    Prove that there are at least positive composite numbers with digits, having a digit equal to and all the rest equal to .
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    Quote Originally Posted by kira View Post
    Prove that there are at least positive composite numbers with digits, having a digit equal to and all the rest equal to .
    This number has to have 2005 digits equal to 1. Notice that 2005 = 333\times6+7. So the number consists of 333 batches consisting of the digits 111111, together with one remaining batch that consists of seven 1s with a 7 somewhere between the 1s.

    Next, notice that 111111 is divisible by both 7 and 13. Also, 11171111 is divisible by 7, and 11111711 is divisible by 13. So if the exceptional batch (the one containing the 7) has the 7 either between the third and fourth 1, or between the fifth and sixth 1, then the entire 2006-digit number will be composite.

    So there are (at least) two composite numbers for each possible placing of the exceptional batch. But there are 334 such placings: the exceptional batch could come at the beginning or the end of the whole number, or between any consecutive pair of 111111s. That seems to give a total of at least 668 composite numbers, two more than was asked for.
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