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interger

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Uchiha Sasuke
28/05/2018 at 03:29
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Given \(f\left(x\right)=\dfrac{2x^2+3x+3}{2x-1}.\) If a and f(a) are both integer we say that f(a) is an element of the set A. What is the sum of all elements of the set A?

interger


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Uchiha Sasuke
08/06/2018 at 02:26
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Given that  \(a^2-b^2=1\). Evaluate \(A=2\left(a^6-b^6\right)-3\left(a^4+b^4\right)\).

interger


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Uchiha Sasuke
11/11/2017 at 20:06
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The greatest value of integer x satisfying \(\left|x-1\right|+\left|x-3\right|+\left|x-5\right|+\left|x-7\right|=8\)

interger

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    Nguyễn Đặng Hoàng Phúc 14/11/2017 at 13:38

    We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

    \(\ge\)|x-1+x-3+5-x+7-x|=|8|=8

    Mark ''='' occurs when \(\left(x-1\right)\left(x-3\right)\left(5-x\right)\left(7-x\right)\ge0\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)\ge0\)

    Case 1 :x-1 , x-3 , x-5 , x-7 \(\ge0\) so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

    Case 2 :x-1 , x-3\(\ge\)0 ; x-5 , x-7\(\le0\)\(\Rightarrow x-3\ge0;x-5\le0\)\(\Rightarrow5\ge x\ge3\)

    Case 3 :x-1 , x-5 \(\ge\)0 ; x-3 , x-7\(\le\)0\(\Rightarrow x-5\ge0;x-3\le0\)\(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\)(not satisfy)

    Case 4 :x-1 , x-7 \(\ge0\) ; x-3 , x-5\(\le\)0\(\Rightarrow\left\{{}\begin{matrix}x\ge7\\x\le3\end{matrix}\right.\)(not satisfy)

    Case 5 :x-1 , x-3 , x-5 , x-7\(\le\)0\(\Rightarrow x\le1\)

    Case 6 :x-1 , x-3\(\le0\) ; x-5 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

    Case 7 :x-1 , x-5\(\le\)0 ; x-3 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

    Case 8 :x-1 , x-7\(\le0\) ; x-3 , x-5\(\ge0\)\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge5\end{matrix}\right.\)(not satisfy)

    Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

    So maxx=5

    Selected by MathYouLike
  • ...
    FA KAKALOTS 09/02/2018 at 21:16

    We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

    ≥

    |x-1+x-3+5-x+7-x|=|8|=8

    Mark ''='' occurs when (x−1)(x−3)(5−x)(7−x)≥0

    ⇒(x−1)(x−3)(x−5)(x−7)≥0

    Case 1 :x-1 , x-3 , x-5 , x-7 ≥0

     so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

    Case 2 :x-1 , x-3≥

    0 ; x-5 , x-7≤0⇒x−3≥0;x−5≤0⇒5≥x≥3

    Case 3 :x-1 , x-5 ≥

    0 ; x-3 , x-7≤0⇒x−5≥0;x−3≤0⇒{x≥5x≤3

    (not satisfy)

    Case 4 :x-1 , x-7 ≥0

     ; x-3 , x-5≤0⇒{x≥7x≤3

    (not satisfy)

    Case 5 :x-1 , x-3 , x-5 , x-7≤

    0⇒x≤1

    Case 6 :x-1 , x-3≤0

     ; x-5 , x-7≥0⇒{x≤1x≥7

    (not satisfy)

    Case 7 :x-1 , x-5≤

    0 ; x-3 , x-7≥0⇒{x≤1x≥7

    (not satisfy)

    Case 8 :x-1 , x-7≤0

     ; x-3 , x-5≥0⇒{x≤1x≥5

    (not satisfy)

    Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

    So maxx=5

  • ...
    Phước Lợi 26/11/2017 at 21:36

    Các bạn lập bảng xét dấu: như vậy chỉ cần xét 5 trường hợp thôi:

    1)  \(x\le1\) thì các biểu thức trong trị tuyệt đối đều âm ta có  VT=(1-x)+(3-x)+(5-x)+(7-x)=8

    ta giải tìm x

    Tương tự cho 4 trường hợp còn lại. undefined


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Uchiha Sasuke
11/11/2017 at 20:03
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Find the positive integer x such that
\(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)

interger

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    Dao Trong Luan Coordinator 11/11/2017 at 20:56

    For \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)

    => \(x^2-1;x^2-4;x^2-7;x^2-10\) have a positive and 3 negative or a nagative and 3 positive

    * If have a positive and 3 negative:

    => x2 - 1 > 0 > x2 - 4 > x2 - 7 > x2 - 10

    => x2 - 1 > 0 and x2 - 4 < 0

    => 1 < x2 < 4 => x2 can equal 2 or 3.

     But x is integer so x2 can't equal 2 or 3

    * If have a negative and 3 positive:

    => x2 - 1 > x2 - 4 > x2 - 7 > 0 > x2 - 10

    => x2 - 7 > 0 and x2 - 10 < 0

    => 7 < x2 < 10

    => x2 = 9 because x is integer

    => x = \(\pm3\)

    But x is the positive integer => x = 3

    Selected by MathYouLike
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    Nguyễn Hưng Phát 12/11/2017 at 22:16

    Equivalent disequations:\(\left(x^4-11x^2+10\right)\left(x^4-11x^2+28\right)< 0\)

    Put \(x^4-11x^2+10=t\)

    We have:t(t+18)<0

    TH1:\(\left\{{}\begin{matrix}t>0\\t+18< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t>0\\t< -18< 0\end{matrix}\right.\) (loại)

    TH2:\(\left\{{}\begin{matrix}t< 0\\t+18>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t< 0\\t>-18\end{matrix}\right.\)\(\Rightarrow t\in\left\{-17,-16,...........,-1\right\}\)

    It's easy here, you try it yourself......

  • ...
    Dao Trong Luan Coordinator 11/11/2017 at 21:06

    For \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-1\right)< 0\)

    => In this four numbers, have:

    \(\left[{}\begin{matrix}\text{a positive number and 3 negative numbers}\\\text{a negative number and 3 positive numbers}\end{matrix}\right.\)

    * If have a positive number and 3 negative numbers:

    => x2-1 > 0 > x2-4 > x2-7 > x2-10

    => x2 - 1 > 0 and x2 < 4

    => 1 < x2 < 4 => No number is satisfy

    * If have a negative number and 3 positive numbers:

    => x2-1 > x2-4 > x2-7 > 0 > x2-10

    => x2-7 > 0 and x2-10 < 0

    => 7 < x2 < 10

    => x2 = 9 because x is a positive integer <=> x2 is a positive integer too.

    => x = 3 because x is a positive integer


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Carter
10/04/2017 at 08:41
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(1995 IMO shortlisted problem)

Let k be a positive integer. Prove that there are infinitely many perfect squares of the form n2k -7; where n is a positive integer.

 

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Carter
10/04/2017 at 08:39
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Let a;b;c be integers such that \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=3\). Prove that abc is the cube of an integer.

intergernumber

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    John 10/04/2017 at 15:24

    Without loss of generality, we may assume gcd(a,b,c) = 1.

    (otherwise, if d=gcd(a,b,c) the for a'=a/d, b'=b/d, c'=c/d, the equation still holds for a', b', c' and a'b'c' is still a cube if only if abc is a cube).

    We multiply equation by abc, we have:

      \(a^2c+b^2a+c^2b=3abc\)(*)

    if \(abc=\pm1\), the problem is solved.

    Otherwise, let p be a prime divisor of abc. Since gcd(a,b,c)=1, the (*) implies that p divides exactly two of a, b,c. By symetry, we may assume p divides a, b but not c. Suppose that the lagest powers of p dividing a, b are m, n, respecively.

    If n < 2m, then \(n+1\le2m\)  and \(p^{n+1}\)| \(a^2c,b^2c,3abc\). Hence \(p^{n+1}\)|\(c^2b\), forcing \(\)\(p\)|\(c\) (a contradiction). If n > 2m, then \(n\ge2m+1\) and \(p^{2m+1}\)|\(c^2b,b^2a,3abc\). Hence \(p^{2m+1}\)|\(a^2c\), forcing \(p\)|\(c\) (a contradicton). Therefore n = 2m and \(abc=\Pi p^{3m}\), \(p\)|\(abc\), is a cube.

    Carter selected this answer.

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Carter
10/04/2017 at 08:47
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Find all integer solutions of the system of equations x + y + z =3 and x3 + y3 + z3 =3 .

intergerequations

  • ...
    John 10/04/2017 at 15:29

    We have:

    \(\left(x+y+z\right)^3-\left(x^3+y^3+z^3\right)=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

    \(\Rightarrow8=\left(3-z\right)\left(3-x\right)\left(3-y\right)\)   (1)

    Since \(\left(3-x\right)+\left(3-y\right)+\left(3-z\right)=9-\left(x+y+z\right)=9-3=6\)  (2)

    From (1) and (2) we infer \(\left(x,y,z\right)\) would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).

    Carter selected this answer.
  • ...
    FA KAKALOTS 08/02/2018 at 22:03

    We have:

    (x+y+z)3−(x3+y3+z3)=3(x+y)(y+z)(z+x)

    ⇒8=(3−z)(3−x)(3−y)

       (1)

    Since (3−x)+(3−y)+(3−z)=9−(x+y+z)=9−3=6

      (2)

    From (1) and (2) we infer (x,y,z)

     would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).


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