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Math Help - Quick Calculation

  1. #1
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    Quick Calculation

    Evaluate \sqrt{2007\times2008\times2009\times2010-1}-2008^2 without a calculator.
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    n(n+1)(n+2)(n+3) = [n(n+3)][(n+1)(n+2)] = [n^2+3n][n^2 + 3n + 2] = [(n^2+3n+1)-1][(n^2+3n+1)+1]  = (n^2 + 3n+1)^2 - 1.

    So are you sure it is n(n+1)(n+2)(n+3) - 1? If it was n(n+1)(n+2)(n+3) + 1 it would turn into a square. As the above squares?
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  3. #3
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    Hello, math sucks!

    Could there be a typo?
    If the first minus-sign is a plus, there's a neat answer . . .


    Evaluate \sqrt{2007\times2008\times2009\times2010 \:{\color{red}+}\:1}-2008^2 without a calculator.
    Let x = 2008

    Under the radical, we have: . (x-1)x(x+1)(x+2) +1

    . . . . . . =\;[(x-1)(x+2)]\,[x(x+1)] + 1

    . . . . . . = \;[x^2 + x - 2]\,[x^2 + x] + 1

    . . . . . . =\;[x^2 + x - 1 - 1]\,[x^2+x-1 + 1] + 1

    . . . . . . =\;\underbrace{[(x^2+x-1) - 1]\,[(x^2+x-1) + 1]}_{(a-b)(a+b)\:=\:a^2-b^2} + 1

    . . . . . . =\;(x^2+x-1)^2 - 1^2 + 1

    . . . . . . =\;(x^2+x-1)^2


    The problem becomes: . \sqrt{x^2+x-1)^2} - x^2 \;=\;x^2+x-1-x^2 \;=\;x-1


    Therefore, the value is: . 2008 - 1 \;=\;{\color{blue}2007}

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  4. #4
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    Yes, there should be a + instead of -. Apologies for the mistake and thank you all for helping.
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, math sucks!

    Could there be a typo?
    If the first minus-sign is a plus, there's a neat answer . . .
    Might I point out that even if there is a minus, the answer stays 2007. I did it on my calculator just now.
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  6. #6
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    Quote Originally Posted by janvdl View Post
    Might I point out that even if there is a minus, the answer stays 2007. I did it on my calculator just now.
    How can it possible be the same. You are subtracting a number. It should only get smaller.
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How can it possible be the same. You are subtracting a number. It should only get smaller.
    I promise you the answer stays 2007, whether it's a plus or minus. Do it on a calculator and see.
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by janvdl View Post
    I promise you the answer stays 2007, whether it's a plus or minus. Do it on a calculator and see.
    For some reason it seems my calculator ignores any constants I put under the square root. Ignore my last posts. There must be a problem with my calculator.


    EDIT: Ah I think I know what it is. The product under the square root is a very large number and subtracting or adding 1 does not change it much. Therefore when finding the root, one gets the same answer.
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  9. #9
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    Quote Originally Posted by janvdl View Post
    I promise you the answer stays 2007, whether it's a plus or minus. Do it on a calculator and see.
    I do not have to see anything. If \sqrt{n} happens to be an integer then \sqrt{n-2} shall not be an integer (where n\geq 2). So there is no way how you can get the same answer (which is an integer).

    It is time to buy an new calculator.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    Might I point out that even if there is a minus, the answer stays 2007. I did it on my calculator just now.
    You aren't carrying enough digits on your calculator. I get 2006.99999980 using the - sign.

    -Dan
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  11. #11
    Bar0n janvdl's Avatar
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    Quote Originally Posted by topsquark View Post
    You aren't carrying enough digits on your calculator. I get 2006.99999980 using the - sign.

    -Dan
    Realised that just now. You will notice I edited my post.
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