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Math Help - Problem Solving #1

  1. #1
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    Problem Solving #1

    Can an integer other than 1 with all its digits equal to 1 be the square of an
    integer? Prove or disprove.



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  2. #2
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    Well, 100 = 10^2, 10000 = 100^2,.....
    S'that what you mean?
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  3. #3
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    Quote Originally Posted by matgrl View Post
    Can an integer other than 1 with all its digits equal to 1 be the square of an

    integer? Prove or disprove.


    So you're asking if any number of the form 11...1 can be a perfect square?
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  4. #4
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    No - it can't be.

    Hint - Look at the digit at the tenth place for such a number. It has to be even - which gives us a contradiction.
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  5. #5
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    I guess....there should be some sort of way to solve this.
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  6. #6
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    Could you further explain this idea...I am not sure I understand.
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  7. #7
    Senior Member roninpro's Avatar
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    Quote Originally Posted by aman_cc View Post
    No - it can't be.

    Hint - Look at the digit at the tenth place for such a number. It has to be even - which gives us a contradiction.
    Could you clarify this? The way I am reading it, it is not correct. For example, 16 is a perfect square, yet it has a 1 in the tens place.


    One way to look at the problem is to consider all of the squares modulo 100. It's a little computationally intensive, but it will do the job. I computed the list in Mathematica:

    {0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 21, 44, 69, 96, 25, 56, 89, 24, 61, 0, 41, 84, 29, 76, 25, 76, 29, 84, 41, 0, 61, 24, 89, 56, 25, 96, 69, 44, 21, 0, 81, 64, 49, 36, 25, 16, 9, 4, 1, 0}

    We can see that 11 does not appear on the list. Therefore, any number of the form 11...1 (other than 1) cannot be a perfect square.
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    "Notice" that: 111111111111...111111111 is odd.

    Now prove that 1111111...111111111 - 1 is not divided by 8 (why it will prove that any number of the from 11...1 (other than 1) cannot be a perfect square?)
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  9. #9
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    What I meant is this.
    For any perfect square to end in 1, say x = y^2. 'y' can only by of the form 10k + 1 or 10k - 1. This is easy to verify.
    For either of these cases - the tenth place in 'x' will be even. This can be proved easily.

    So, my claim is any perfect square ending in 1 will have an even number in it's tens place. And, number of the form 1111....111 violate that and hence can't be a perfect square.
    Last edited by aman_cc; November 3rd 2010 at 09:21 AM.
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  10. #10
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    @Also sprach Zarathustra - A very nice way to prove this. Thanks
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  11. #11
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    Aman_cc I understand that I need to look at the tenths place...but would I do something like this:

    9^2 = 81
    29 ^2 = 841
    11 ^2 = 121
    21^2 = 441

    I thought the question was asking if all numbers such as 1,11,111,1111,11111 be a square of an integer.


    Could you possibly show me some examples of your thinking because I think you might be right.
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  12. #12
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    I don't think you got my argument right.

    Step 1: Prove that any perfect square with digit 1 at it's one's place will have an even digit at it's ten's place.

    Can you do this using the hints I provided?

    Step 2: A number of the form 11...11111 is clearly a voilation of the argument above, hence, can't be a perfect square.

    This completes the proof. But I think the method by Also sprach Zarathustra is pretty neat/better
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