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Math Help - not integer number

  1. #1
    Super Member dhiab's Avatar
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    Exclamation not integer number

    Prove that number \sqrt{20092010201120122013} is not
    integer number .
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  2. #2
    Super Member Bacterius's Avatar
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    A perfect square never ends in 3, therefore 20092010201120122013 is not a perfect square, therefore \sqrt{20092010201120122013} is not an integer.

    Come on Dhiab, that was too easy, where are the headcracking problems you used to post

    PS : the proof for my original statement is quite trivial, a quick Google search will have it done.
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  3. #3
    Super Member dhiab's Avatar
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    Hello : have the demonstration 'A perfect square never ends in '(Come on Dhiab, that was too easy)
    Thanks
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  4. #4
    Super Member Bacterius's Avatar
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    Fine, I'll give it

    Let n be any integer number. n^2 is its square. Let us look at the last digit of n^2, using arithmetic modulo 10. The following table enumerates all possible cases :

    \begin{array}{|l|l|}<br />
\hline n \mod 10 &n^2 \mod 10 \\<br />
\hline 0 &0 \\<br />
1 &1 \\<br />
2 &4 \\<br />
3 &9 \\<br />
4 &6 \\<br />
5 &5 \\<br />
6 &6<br />
\end{array}
    \begin{array}{|l|l|}<br />
7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\<br />
8 &4 \\<br />
9 &1 \\<br />
\hline<br />
\end{array}

    Sorry, I had to cut the table midway, it was too big for MHF.

    This shows that a perfect square may only end by the digits 0, 1, 4, 5, 6 and 9. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

    \boxed{\mathrm{QED}}

    This implies that if a number terminates by 3, it cannot be a perfect square.
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  5. #5
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    Hello, dhiab!

    Prove that number \sqrt{20092010201120122013} is not an integer.
    I assume that the author forgot (or was unaware of) the Final Digit Rule for integer squares.

    If that is true, that number has a special rhythm to it: . 2009\,2010\,2011\,2012\,2013

    We have: . N \;=\;10^{16}a + 10^{12}(a+1) + 10^8(a+2) + 10^4(a+3) + (a+4)\;\text{ where }a = 2009

    And I suspect that we are expected to prove algebraically that N is not a square.

    Any ideas?

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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Bacterius View Post
    Fine, I'll give it

    Let n be any integer number. n^2 is its square. Let us look at the last digit of n^2, using arithmetic modulo 10. The following table enumerates all possible cases :

    \begin{array}{|l|l|}<br />
\hline n \mod 10 &n^2 \mod 10 \\<br />
\hline 0 &0 \\<br />
1 &1 \\<br />
2 &4 \\<br />
3 &9 \\<br />
4 &6 \\<br />
5 &5 \\<br />
6 &6<br />
\end{array}
    \begin{array}{|l|l|}<br />
7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\<br />
8 &4 \\<br />
9 &1 \\<br />
\hline<br />
\end{array}

    Sorry, I had to cut the table midway, it was too big for MHF.

    This shows that a perfect square may only end by the digits 0, 1, 4, 5, 6 and 9. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

    \boxed{\mathrm{QED}}

    This implies that if a number terminates by 3, it cannot be a perfect square.
    You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
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  7. #7
    Super Member dhiab's Avatar
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    Originally Posted by :Swlabr
    "You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
    Hello :
    you know that the rest of division ,by 10, a integer number is first digit.
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  8. #8
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    And there's always the brute force method:
    \sqrt{20092010201120122013}=4482411203.93...

    - Hollywood
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by dhiab View Post
    Originally Posted by :Swlabr
    "You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
    Hello :
    you know that the rest of division ,by 10, a integer number is first digit.
    a \equiv 3 \text{ mod } 10 \Rightarrow a \equiv 3 \text{ mod } 5.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by hollywood View Post
    And there's always the brute force method:
    \sqrt{20092010201120122013}=4482411203.93...

    - Hollywood
    Going along these lines you could see that  4482411203^2<20092010201120122013 and  4482411204^2>20092010201120122013 .
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  11. #11
    Newbie Xitami's Avatar
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    Code:
       10001000100010001*2009 + 
           1000200030004 =
    20092010201120122013 = 7*2870287171588588859
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