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Math Help - How many triplets

  1. #1
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    How many triplets

    If we have X,Y,Z,8 are digits for a number that have 4 digits and we arrange them in two ways:
    8 x y z , z y x 8 if x>y>z and 8 x y z-z y x 8=7623 how many triplets (x ,y , z) are possible ?
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  2. #2
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    Re: How many triplets

    I can't think of a shortcut method to solve this -- but working through the inequalities I can only find one triplet.

    Since x>y>z and x\leq9, notice that z\leq7 and y\leq8. So when looking at:
    8 x y z
    \underline{-z y x 8}
    7623

    (sorry, I can't get the numbers above to line up) you'll have to borrow when you begin your subtraction from the right hand column. Remember, when you borrow, you add 10 to z. So right from the start, you know that z+10 - 8 = 3 and so z=1.

    Then it gets a little tricky. You're in the second to right-most column, i.e. y - x = 2. But you borrowed in your first step, so this becomes y-1 -x = 2. Since y \leq x, surely y-1 \leq x. So you'll have to borrow to carry out this subtraction. When you borrow, you essentially add ten, thus  y +10 -1 -x = 2 and so x=y+7.

    At this point, you can deduce that x=9 and y=2. This is the only possibilty, since if x\leq8, we have y\leq1. But this cannot be, since z=1 and we know that y is larger than that.
    Last edited by jsndacruz; April 16th 2012 at 10:18 AM. Reason: Latex stuff
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  3. #3
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    Re: How many triplets

    Hello, Mhmh96!

    \text{We have }x,y,z,8\text{ are digits for a 4-digit number and we arrange them in two ways: }8 x y z\text{ and }z y x 8.
    \text{If }x>y>z\text{ and }8 x y z - z y x 8\:=\:7623\text{, how many triplets }(x ,y , z)\text{ are possible?}

    We have:

    . . \begin{array}{ccccc}& 8 & x & y & z \\ - & z & y & x & 8 \\ \hline & 7 & 6 & 2 & 3\end{array}


    We see immediately that z = 1

    . . \begin{array}{ccccc}& 8 & x & y & 1 \\ - & 1 & y & x & 8 \\ \hline & 7 & 6 & 2 & 3\end{array}


    In the second column, we see that: x \:=\:y+7
    . . Think about it!


    Hence, we have: . (x,y,z) \;=\;(9,2,1),\;(8,1,1),\;(7,0,1)

    The last two triplets violate the stipulation: x > y > z.


    Therefore, the answer is one triplet: . (9,2,1)

    Thanks from jsndacruz
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  4. #4
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    Re: How many triplets

    Thank you jsndacruz ,Soroban !
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