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Math Help - SAT problem

  1. #1
    klg
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    SAT problem

    the question is:
    A four-digit integer, WXYZ, in which W,X,Y and Z each represent a different digit, is formed according to the following rules.
    1. x=w+y+z
    2. w=Y+1
    3. Z=W-5
    what is the four-digit integer?

    I have tried every system of equation and substitution I can think of!
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  2. #2
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    Re: SAT problem

    Obviously, you have four variables and three equations, so you will not be able to find a unique solution using regular methods for solving systems of equations. You have to use the facts that digits are integers between 0 and 9.

    We have x = 3(y - 1) <= 9. On the other hand, z = y - 4 >= 0, so y >= 4. This gives a unique value for y, from which other digits can be found.

    In mathematics (as well as in physics and programming), it's not good to denote the same variable by a lower- and uppercase letters.
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  3. #3
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    Re: SAT problem

    Quote Originally Posted by klg View Post
    the question is:
    A four-digit integer, WXYZ, in which W,X,Y and Z each represent a different digit, is formed according to the following rules.
    1. x=w+y+z
    2. w=Y+1
    3. Z=W-5
    what is the four-digit integer?

    I have tried every system of equation and substitution I can think of!
    w = y+1
    z = w-5 = (y+1) - 5 = y-4 ... this says y > 4

    x = w+y+z = 3y-3 using the above equations

    x = 3y-3 implies that y has to be one of {1,2,3,4} ... look two lines up for the additional restriction.


    you should be able to figure it out from here ...
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  4. #4
    klg
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    Re: SAT problem

    Got it.
    Thanks!
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  5. #5
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    Re: SAT problem

    Hello, klg!

    This problem doesn't require fancy algebra . . .


    A four-digit integer, WXYZ, in which W,X,Y and Z each represent a different digit,
    is formed according to the following rules.

    . . [1]\;X\,=\,W+Y+Z
    . . [2]\;W\,=\,Y+1
    . . [3]\;Z\,=\,W-5

    What is the four-digit integer?

    From [2]: . Y \,=\,W-1
    From [3]: . Z \,=\,W-5

    Substitute into [1]: . X \:=\:W + (W-1) + (W-5) \quad\Rightarrow\quad X \:=\:3W - 6


    We have: . \begin{Bmatrix}(a) & W &=& W \\ (b) & X &=& 3W - 6 \\ (c) & Y &=& W-1 \\ (d) &Z &=& W - 5 \end{Bmatrix}


    Since X \le 9, we have: . (b)\;3W-6 \,\le\,9 \quad\Rightarrow\quad W \,\le\,5

    Since Z \ge 0, we have: . (d)\;W-5 \,\ge\,0 \quad\Rightarrow\quad W \,\ge\,5

    Therefore: . \begin{Bmatrix}W &=& 5 \\ X &=& 9 \\ Y &=& 4 \\ Z &=& 0 \end{Bmatrix}
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  6. #6
    klg
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    Re: SAT problem

    That is WONDERFULLY simple, and seems so clear now!
    Thank you so much!
    klg
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