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Math Help - Solve for "X" when added to the Square of Number

  1. #1
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    Solve for "X" when added to the Square of Number

    The positive integer X is divisible by 25. If [sqrt]X (square root of X) is greater than 25, what integer between 22 and 26, inclusive, could be the value of X/25?


    I guessed the answer to be 26. The key I have says the answer = 26. I just can't figure out how really?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by thecapaccino View Post
    The positive integer X is divisible by 25. If [sqrt]X (square root of X) is greater than 25, what integer between 22 and 26, inclusive, could be the value of X/25?


    I guessed the answer to be 26. The key I have says the answer = 26. I just can't figure out how really?
    X=25n for some integer n.

    sqrt(X) = 5 sqrt(n)>25, so sqrt(n)>5 and so n>25.

    But n=X/25 and we have seen that n>25, and the only option in the given
    range is n=26.

    RonL
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